RD Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Math Chapter 10 Differentiation are provided here with simple step-by-step explanations. These solutions for Differentiation are extremely popular among class 12 Commerce students for Math Differentiation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2020 Book of class 12 Commerce Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2020 Solutions. All RD Sharma XII Vol 1 2020 Solutions for class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 10.103:

Question 1:

Find dydx, when

x=at2 and y=2 at

Answer:

We have,  x=at2 and  y=2atdxdt=2at and  dydt=2a dydx=dydtdxdt=2a2at=1t

Page No 10.103:

Question 2:

Find dydx, when

x=a θ+sin θ and y=a 1-cos θ

Answer:

We have, x=aθ+sinθ and y=a1-cosθ

dxdθ=a1+cosθ and dydθ=a sinθ    

dydx=dydθdxdθ=a sinθa1+cosθ=2 sinθ2cosθ22cos2θ=tanθ2

Page No 10.103:

Question 3:

Find dydx, when

x=a cos θ and y=b sin θ

Answer:

We have, x=a cosθ and y=b sinθ dxdθ=-a sinθ and dydθ=b cosθdydx=dydθdxdθ=b cosθ-a sinθ=-bacotθ

Page No 10.103:

Question 4:

Find dydx, when

x=aeθ sin θ-cos θ, y=aeθ sin θ+cos θ

Answer:

We have,  x=aeθsinθ-cosθ and y=aeθsinθ+cosθ

dxdθ=aeθddθsinθ-cosθ+sinθ-cosθddθeθ  and dydθ=aeθddθsinθ+cosθ+sinθ+cosθddθeθdxdθ=aeθcosθ+sinθ+sinθ-cosθeθ and dydθ=aeθcosθ-sinθ+sinθ+cosθeθdxdθ=a2eθ sinθ and dydθ=a2eθ cosθ               

dydθdxdθ=a2eθcosθa2eθsinθ=cotθ

Page No 10.103:

Question 5:

Find dydx, when

x=b sin2 θ and y=a cos2 θ

Answer:

We have, x=b sin2θ and y=a cos2θdxdθ=ddθb sin2θ=2b sinθcosθand, dydθ=ddθa cos2θ=-2a cosθsinθ   dydx=dydθdxdθ=-2a cosθsinθ 2b sinθcosθ=-ab

Page No 10.103:

Question 6:

Find dydx, when

x=a 1-cos θ and y=a θ+sin θ at θ=π2

Answer:

We have, x=a1-cosθ  and  y=aθ+sinθ dxdθ=ddθa1-cosθ=asinθand dydθ=ddθaθ+sinθ=a1+cosθ  dydxθ=π2=dydθdxdθθ=π2=a1+cosθasin θθ=π2=a1+0a=1

Page No 10.103:

Question 7:

Find dydx, when

x=et+e-t2 and y=et-e-t2

Answer:

We have, x=et+e-t2 and y=et-e-t2

dxdt=12ddtet+ddte-t and  dydt=12ddtet-ddte-tdxdt=12et+e-tddt-t and dydt=12et-e-tddte-tdxdt=12et-e-t=y and dydt=12et+e-t=x       dydt=dydtdxdt=xy

Page No 10.103:

Question 8:

Find dydx, when

x=3 at1+t2, and y=3 at21+t2

Answer:

We have, x=3at1+t2
Differentiating with respect to t,
dxdt=1+t2ddt3at-3atddt1+t21+t22                     using quotient ruledxdt=1+t23a-3at2t1+t22dxdt=3a+3at2-6at21+t22dxdt=3a-3at21-t22dxdt=3a1-t21+t22                           ...iand, y=3at21+t2
Differentiating it with respect to t,

dxdt=1+t2ddt3at2-3at2ddt1+t21+t22                     using quotient ruledxdt=1+t26at-3at22t1+t22dxdt=6at+6at3-6at31+t22dxdt=6at1+t22                         ...iiDividing equation ii by i,dydtdxdt=6at1+t22×1+t223a1-t2=2t1-t2

Page No 10.103:

Question 9:

Find dydx, when

x=a cos θ + θ sin θ and y=a sin θ-θ cos θ

Answer:

We have, x=acosθ+θ sinθ and y=asinθ-θ cosθdxdθ =addθcosθ+ddθθ sinθ and dydθ=addθsinθ-ddθθ cosθdxdθ =a-sinθ+θddθsinθ+sinθddθθ and dydθ=acosθ-θddθcosθ+cosθddθθ dxdθ =a-sinθ+θ cosθ and dydθ=acosθ+θ sinθ-cosθdxdθ  =aθ cosθ and dydθ=aθ sinθ dydx=dydθdxdθ=aθ sinθaθ cosθ =tanθ  

Page No 10.103:

Question 10:

Find dydx, when

x=eθ θ+1θ and y=e-θ θ-1θ

Answer:

We have, x=eθθ+1θ
Differentiating it with respect to θ,
dxdθ=eθddθθ+1θ+θ+1θddθeθ                       using product ruledxdθ=eθ1-1θ2+θ2+1θeθdxdθ=eθ1-1θ2+θ2+1θdxdθ=eθθ2-1+θ3+θθ2dxdθ=eθθ3+θ2+θ-1θ2                       ...iand,       y=eθθ-1θ 
Differentiating it with respect to θ using chain rule,

dydθ=e-θddθθ-1θ+θ-1θddθe-θ                       using product ruledydθ=e-θ1+1θ2+θ-1θeθddθ-θdydθ=e-θ1+1θ2+θ-1θe-θ-1dydθ=e-θ1+1θ2-θ+1θdydθ=e-θθ2+1-θ3+θθ2dydθ=e-θ-θ3+θ2+θ+1θ2                     ...iiDividing equation ii by i,dydθdxdθ=e-θθ2-θ3+θ+1θ2×θ2eθθ3+θ2+θ-1        =e-2θθ2-θ3+θ+1θ3+θ2+θ-1 

Page No 10.103:

Question 11:

Find dydx, when

x=2 t1+t2 and y=1-t21+t2

Answer:

We have, x=2t1+t2

dxdt=1+t2ddt2t-2tddt1+t21+t22                          using quotient ruledxdt=1+t22-2t2t1+t22dxdt=2+2t2-4t21+t22dxdt=2-2t21+t22                              ...iand,y=1-t21+t2

dydt=1+t2ddt1-t2-1-t2ddt1+t21+t22dydt=1+t2-2t-1-t22t1+t22dydt=-4t1+t22                          ...iiDividing equation ii by i, we get,dydtdxdt=-4t1+t22×1+t2221-t2dydx=-2t1-t2dydx=-xy                         xy =2t1+t2×1+t21-t2=2t1-t2

Page No 10.103:

Question 12:

Find dydx, when

x=cos-1 11+t2 and y=sin-1 t1+t2, t  R

Answer:

We have, x=cos-111+t2

dxdt=-11-11+t22ddt11+t2dxdt=-11-11+t2-121+t232ddt1+t2dxdt=1+t2121+t2-1×121+t2322tdxdt=tt2×1+t2dxdt=11+t2                                   ...iNow, y=sin-111+t2

dydt=11-11+t22ddt11+t2dydt=11-11+t2-121+t232ddt1+t2dxdt=1+t2121+t2-1×-121+t2322tdxdt=-12t2×1+t22tdxdt=-11+t2                                   ...iiDividing equation ii by i,dydtdxdt=11+t2×1+t2-1dydx=-1

Page No 10.103:

Question 13:

Find dydx, when

x=1-t21+t2 and y=2 t1+t2

Answer:

We have, y=2t1+t2

dydt=1+t2ddt2t-2tddt1+t21+t22                          using quotient ruledydt=1+t22-2t2t1+t22dydt=2+2t2-4t21+t22dydt=2-2t21+t22                              ...iand,x=1-t21+t2

dxdt=1+t2ddt1-t2-1-t2ddt1+t21+t22dxdt=1+t2-2t-1-t22t1+t22dxdt=-4t1+t22                          ...iiDividing equation i by ii, we get,dydtdxdt=21-t21+t22×1+t22-4tdydx=21-t2-4tdydx=t2-12t                      

Page No 10.103:

Question 14:

If x=2 cos θ-cos 2 θ and y=2 sin θ-sin 2 θ, prove that dydx=tan 3 θ2

Answer:

We have, x=2 cosθ- cos2θ

dxdθ=2-sinθ--sin2θddθ2θdxdθ=-2sinθ+2 sin2θdxdθ=2sin2θ-sinθ              ...iand,y=2 sinθ-sin2θ

dydθ=2 cosθ-cos2θddθ2θdydθ=2 cosθ-cos2θ2dydθ=2 cosθ-2 cos2θdydθ=2cosθ-cos2θ              ...iiDividing equation ii by equation i,dydθdxdθ=2cosθ-cos2θ2sin2θ-sinθdydx=cosθ-cos2θsin2θ-sinθdydx=-2sinθ+2θ2sinθ-2θ22cos2θ+θ2sin2θ-θ2                        sinA-sinB=2 cosA+B2sinA-B2and cosA-cosB=-2sinA+B2sinA-B2dydx=-sin3θ2sin-θ2cos3θ2sinθ2dydx=-sin3θ2-sinθ2cos3θ2sinθ2dydx=sin3θ2cos3θ2dydx=tan3θ2

Page No 10.103:

Question 15:

If x=ecos 2 t and y=esin 2 t, prove that dydx=-y log xx log y

Answer:

We have, x=ecos2t and y=esin2t

dxdt=ddtecos2t and dydt=ddtesin2tdxdt=ecos2tddtcos2t and dydt=esin2tddtsin2tdxdt=ecos2t-sin2tddt2t and dydt=esin2tcos2tddt2t            dxdt=-2sin 2tecos2t anddydt=2cos2tesin2t           dydtdxdt=2 cos2tesin2t-2sin2tecos2tdydx=-y logxx logy                                     x=ecos2tlogx=cos2ty=esin2tlogy=sin 2t                    

 

Page No 10.103:

Question 16:

If x=cos t and y=sin t, prove that dydx=13 at t=2 π3

Answer:

We have, x=cost and y=sin t

dxdt=ddtcos t and dydt=ddtsin tdxdt=-sin t and dydt=cos t            dydtdxdt=cos t-sin t=-cot t    Now,  dydxt=2π3=-cot 2π3=13                                                    

Page No 10.103:

Question 17:

If x=at+1t and y=at-1t, prove that dydx=xy

Answer:

We have,  x=at+1t and y=at-1t

dxdt=addtt+1t and dydt=addtt-1tdxdt=a1-1t2 and dydt=a1+1t2  dxdt=at2-1t2 and dydt=at2+1t2                 dydx = dydtdxdt=at2+1t2×t2at2-1dydx =at2+1t×tat2-1dydx =at+1t×1at-1t dydx=xy                                                                               

Page No 10.103:

Question 18:

If x=sin-1 2 t1+t2 and y=tan-1 2 t1-t2,-1<t<1, prove that dydx=1

Answer:

We have, x=sin-12t1+t2Put t=tanθ-1<tanθ<1-π4<θ<π4-π2<2θ<π2x=sin-12 tanθ1+tan2θ x=sin-1sin2θx=2θ                                -π2<2θ<π2x=2tan-1t                   t=sinθ

dxdt=21+t2           ...iNow, y=tan-12t1-t2put t=tanθ  y=tan-12 tanθ1-tan2θ  y=tan-1tan 2θ               y=2θ                                     -π2<2θ<π2 y=2 tan-1t                              t=tanθ    

dydt=21+t2        ...iiDividing equation ii by i,dydtdxdt=21+t2×1+t22dydx=1

Page No 10.103:

Question 19:

If x=sin3 tcos 2 t, y=cos3 tcost 2 t, find dydx

Answer:

We have, x=sin3tcos2t  and  y=cos3tcos2tdxdt=ddtsin3tcos2t dxdt=cos2tddtsin3t-sin3tddtcos2tcos2t              Using quotient ruledxdt=cos2t3sin2tddtsint-sin3t×12cos2tddtcos 2tcos2t       dxdt=3cos2tsin2t cost-sin3t2cos2t-2 sin2tcos 2t      dxdt=3cos2t sin2t cost+sin3t sin2tcos2tcos2tNow, dydt=ddtcos3tcos2t       dydt=cos2tddtcos3t-cos3tddtcos2tcos2t              Using quotient rule      dydt=cos2t3cos2tddtcost-cos3t×12cos2tddtcos 2tcos2t       dydt=3cos2tcos2t -sint-cos3t2cos2t-2 sin2tcos 2t      dydt=-3cos2t cos2t sint+cos3t sin2tcos2tcos2tdydx=dydtdxdt=-3cos2t cos2t sint+cos3t sin2tcos2tcos2t×cos2tcos2t3cos2t sin2t cost+sin3t sin2t    dydx=-3cos2t cos2t sint+cos3t sin2t3cos2t sin2t cost+sin3t sin2t    dydx=sint cost-3cos2t cost+2cos3tsint cost3cos2t sint+2sin3t    dydx=-32cos2t-1cost+2cos3t31-2sin2tsint+2sin3t                    cos2t=2cos2t-1cos2t=1-2sin2t    dydx=-4cos3t+3cost3sint-4sin3t    dydx=-cos3tsin3t                                   cos3t=4cos3t-3costsin3t=3sint-4sin3tdydx=-cot3t

Page No 10.103:

Question 20:

If x=t+1ta, y=at+1t, find dydx

Answer:

We have, x=t+1ta

dxdt=ddtt+1ta                   dxdt=at+1ta-1ddtt+1tdxdt=at+1ta-11-1t2            ...i             and, y=at+1t

dydt=ddtat+1t         dydt=at+1t× logaddtt+ttdydt=at+1t× loga1-1t2          ...iiDividing equation ii by i,dydtdxdt=at+1t× loga1-1t2  at+1ta-11-1t2dydx=at+1t× logaat+1ta-1

Page No 10.103:

Question 21:

If x=a 1+t21-t2 and y=2t1-t2, find dydx

Answer:

We have, x=a1+t21-t2

dxdt=a1-t2ddt1+t2-1+t2ddt1-t21-t22                    Using quotient ruledxdt=a1-t22t-1+t2-2t1-t22dxdt=a2t-2t3+2t+2t31-t22dxdt=4at1-t22                ...iand, y=2t1-t2

dydt=21-t2ddtt-tddt1-t21-t22                    Using quotient ruledydt=21-t21-t-2t1-t22dydt=21-t2+2t21-t22dydt=21+t21-t22                ...iiDividing equation ii by i,dydtdxdt=21+t21-t22  ×1-t224atdydx=1+t22at



Page No 10.104:

Question 22:

If x=10 t-sin t, y=12 1-cos t, find dydx.

Answer:

We have, x=10t-sint and y=121-costdxdt=ddt10t-sint anddydt=ddt121-costdxdt=10ddtt-sint and dydt=12ddt1-costdxdt=101-cost and dydt=120--sint=12 sintdydx=dydtdxdt=12 sint101-costdydx=12×2sint2cost210×2sin2t2dydx=65cott2

Page No 10.104:

Question 23:

If x=a θ-sin θ and, y=a 1+cos θ, find dydx at θ=π3.

Answer:

We have, x=aθ-sinθ  and  y=a1+cosθ dxdθ=ddθaθ-sinθ and dydθ=ddθa1+cosθdxdθ=a1-cosθ and dydθ=a-sinθ dydx=dydθdxdθ=-a sinθa1-cosθNow, dydxθ=π3=-sinπ31-cosπ3=-321-12=-3

Page No 10.104:

Question 24:

If x=asin2t1+cos2t and y=bcos2t1-cos2t, show that at t=π4, dydx=bat=π4, dydx=ba

Answer:

x=asin2t1+cos2t and y=bcos2t1-cos2tdxdt=2acos2t1+cos2t+2asin2t1-cos2t and dydt=-2bsin2t1-cos2t+2bcos2t1+cos2t dxdt=2acos2t+cos22t+sin2t-sin2tcos2t and dydt=2b-sin2t+sin2tcos2t+cos2t+cos22tdydt=dydtdxdt=-2b-sin2t+sin2tcos2t+cos2t+cos22t2acos2t+cos22t+sin2t-sin2tcos2tdydtt=π4=-2b-sinπ2+sinπ2cosπ2+cosπ2+cos2π22acosπ2+cos2π2+sinπ2-sinπ2cosπ2=ba

Page No 10.104:

Question 25:

If x=cost3-2cos2t, y=sint3-2sin2t find the value of dydx at t=π4

Answer:

x=cost3-2cos2t and y=sint3-2sin2tdxdt=-sint3-2cos2t+cost4costsint and dydt=cost3-2sin2t+sint-4sintcostdxdt=-3sint+6sintcos2t and dydt=3cost-6sin2tcostdxdt=-3sint1-2cos2t and dydt=3cost1-2sin2tdxdt=3sintcos2t and dydt=3costcos2tdydx=dydtdxdt=3costcos2t3sintcos2t=cottNow, dydxt=π4=cotπ4=1

Page No 10.104:

Question 26:

If x=1+logtt2, y=3+2logtt, find dydx

Answer:

x=1+logtt2 and y=3+2logttdxdt=t-2t-2tlogtt4 and dydt=2-3-2logtt2dxdt=-1-2logtt3 and dydt=-1-2logtt2dydx=dydtdxdt=-1-2logtt2-1-2logtt3=t

Page No 10.104:

Question 27:

sinx=2t1+t2, tany=2t1-t2, find dydx

Answer:

sinx=2t1+t2 and tany=2t1-t2x=sin-12t1+t2 and y=tan-12t1-t2x=2tan-1t and y=2tan-1tdxdt=2t1+t2 and dydt=2t1+t2dydx=dydtdxdt=2t1+t22t1+t2=1

Page No 10.104:

Question 28:

Write the derivative of sinx with respect to cosx

Answer:

Let u=sinx and v=cosxdudx=cosx and dvdx=-sinxdudxdvdx=cosx-sinxdudv=-cotx

Page No 10.104:

Question 29:

If x = a (2θ –  sin 2θ) and y = a (1 – cos 2θ), find dydx when θ=π3.

Answer:

Given values are:
x=a2θ-sin2θandy=a1-cos2θ

Applying parametric differentiation

dxdθ= 2a − 2acos2θ

dydθ= 0 + 2asin2θ

dydxdydθ×dθdx=sin2θ1- cos2θ

Now putting the value of θπ3
dydxθ=π3=sin2π31-cos2π3                =321+12                =3232=13
So, dydx is 13 at θ=π3.



Page No 10.112:

Question 1:

Differentiate x2 with respect to x3

Answer:

Let u=x2  and v=x3dudx=2x and dvdx=3x2     dudv=dudxdvdx=2x3x2=23x

Page No 10.112:

Question 2:

Differentiate log (1 + x2) with respect to tan−1x

Answer:

Let u=log1+x2 and v=tan-1x dudx=11+x2ddx1+x2 =2x1+x2 and dvdx=11+x2                   dudv=dudxdvdx=2x1+x2×1+x21=2x
 

Page No 10.112:

Question 3:

Differentiate (log x)x with respect to log x

Answer:

Let u=logxx
Taking log on both sides,
log u=loglogxxlog u=x loglogx           

1ududx=xddxloglogx+loglogxddxx1ududx=x1logxddxlogx+loglogx1dudx=uxlogx1x+log logxdudx=logxx1logx+log logx                   ..iAgain, let v=logxdvdx=1x                     ...iiDividing equation i by ii,we getdudxdvdx=logxx1logx+log logx1xdudv=logxx1+logxlog logxlogx1xdudv=xlogxx-11+logx × log logx

Page No 10.112:

Question 4:

Differentiate sin-1  1-x2 with respect to cos-1 x, if
(i) x  0, 1
(ii) x  -1, 0

Answer:

i Let,    u=sin-11-x2Put    x=cosθ     u=sin-11-cos2θ    u=sin-1sinθ                ...iAnd, v=cos-1x               ...iiNow,   x0,1    cosθ0,1      θ0,π2So, from equation i,   u=θ                   Since, sin-1sinθ=θ if θ-π2,π2u=cos-1x            Since, cosθ=x 

Differentiating it with respect to x,

dudx=-11-x2              ...iiifrom equation ii,v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2           ...ivDividing equation iii by iv,dudxdvdx=-11-x2×1-x2-1dudx=1


ii  Let,    u=sin-11-x2Put    x=cosθ     u=sin-11-cos2θ     u=sin-1sinθ                ...iAnd, v=cos-1x                      ...iiNow,   x-1,0    cosθ-1,0      θπ2,πSo, from equation i,  u=π-θ                   Since, sin-1sinθ=π-θ if θπ2,3π2 u=π-cos-1x            Since, x=cosθ 

Differentiating it with respect to x,

dudx=0--11-x2   dudx=  11-x2           ...iiifrom equation ii,v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2           ...ivDividing equation iii by iv,dudxdvdx=11-x2×1-x2-1dudx=-1

Page No 10.112:

Question 5:

Differentiate sin-1 4x 1-4x2 with respect to 1-4x2, if

(i) x -12 2, 12 2

(ii) x  12 2, 12

(iii) x  -12, -12 2

Answer:

i    Let,   u=sin-14x1-4x2       put    2x=cosθ             u=sin-12×cosθ1-cos2θ             u=sin-12cosθ sinθ              u=sin-1sin 2θ                      ...i     Let,     v=1-4x2                            ...iiHere,         x-122, 122    2x-12, 12  θπ4,3π4So, from equation i,    u=π-2θ                            Since, sin-1sinθ=π-θ ,if θπ2, π u=π-2 cos-12x            Since, 2x=cosθ                   

Differentiating it with respect to x,

dudx=0-2-11-2x2ddx2xdudx=21-4x22dudx=41-4x2                        ...iiifrom equation iidvdx=-4x1-4x2but, x-12,-122dvdx=-4-x1-4-x2dvdx=4x1-4x2                         ...ivDiferentiating equation ii with respect to x,dvdx=121-4x2ddx1-4x2dvdx=121-4x2-8xdvdx=-4x1-4x2                     ...vDividing equation iii by vdudxdvdx=41-4x2 ×1-4x2-4xdudv=-1x


ii    Let,   u=sin-14x1-4x2       put  2x=cosθ                 u=sin-12×cosθ1-cos2θ         u=sin-12cosθ sinθ           u=sin-1sin 2θ                      ...i     Let,     v=1-4x2                         ...iiHere,         x122,12    2x12,1  cosθ12,1  θ0,π4So, from equation i,    u=2θ                            Since, sin-1sinθ=θ ,if θ-π2,π2 u=2 cos-12x            Since, 2x=cosθ                   

Differentiate it with respect to x,

dudx=2-11-2x2ddx2xdudx=-21-4x22dudx=-41-4x2                          ...iiiDiferentiating equation ii with respect to x,dvdx=121-4x2ddx1-4x2dvdx=121-4x2-8xdvdx=-4x1-4x2                            ...ivDividing equation iii by ivdudxdvdx=-41-4x2 ×1-4x2-4xdudv=1x


iii    Let,   u=sin-14x1-4x2       put,  2x=cosθ          u=sin-12×cosθ1-cos2θ          u=sin-12cosθ sinθ           u=sin-1sin 2θ                ...i     Let,     v=1-4x2                   ...iiHere,         x-12,-122    2x-1,-12   θ3π4,πSo, from equation i,    u=π-2θ                            Since, sin-1sinθ=π-θ ,if θπ2,3π2 u=π-2 cos-12x            Since, 2x=cosθ                   


Differentiate it with respect to x,

dudx=0-2-11-2x2ddx2xdudx=21-4x22dudx=41-4x2                         ...iiifrom equation ii,dvdx=-4x1-4x2but, x-12,-122dvdx=-4-x1-4-x2dvdx=4x1-4x2                     ...ivDividing equation iii by ivdudxdvdx=41-4x2 ×1-4x24xdudv=1x

Page No 10.112:

Question 6:

Differentiate tan-1 1+x2-1x with respect to sin-1 2x1+x2, if -1<x<1, x0.

Answer:

Let,     u=tan-11+x2-1xput     x=tanθ       u=tan-11+tan2θ-1tanθ      u=tan-1secθ-1tanθ     u=tan-11-cosθsinθ       u=tan-12sin2θ22sinθ2cosθ2    u=tan-1 tanθ2                     ...iAnd,       v=sin-12x1+x2    v=sin-12tanθ1+tan2θ     v=sin-1sin2θ                      ...iiHere,     -1<x<1 -1<tanθ<1  -π4<θ<π4                       ...A So, from equation i,u=θ2                        Since, tan-1tanθ=θ, if θ-π2,π2 u=12tan-1x             since, x=tanθ

Differentiating it with respect to x,

dudx=1211+x2dudx=121+x2         ...iNow, from equation ii and A,v=2θ                            Since, sin-1sinθ=θ, if θ-π2,π2v=2tan-1x                  Since, x=tanθ

Differentiating it with respect to x,

dvdx=211+x2          ...ivdividing equation iii by iv,dudxdvdx=121+x2×1+x22dudv=14

Page No 10.112:

Question 7:

Differentiate sin-1 2x 1-x2 with respect to sec-1 11-x2, if
(i) x 0, 12

(ii) x 12, 1

Answer:

i Let,    u=sin-12x1-x2     Put    x=sinθ          u=sin-12sinθ1-sin2θ          u=sin-12 sinθ cosθ            u=sin-1sin2θ                     ...iAnd,   Let        v=sec-111-x2            v=sec-111-sin2θ               v=sec-11cosθ              v=sec-1secθ               v=cos-111cosθ                       Since, sec-1x=cos-11x             v=cos-1cosθ                  ...iiHere,           x0,12   sinθ0,12          θ0,π4So, from equation i,    u=2θ                                 Since, sin-1sinθ=θ, if θ-π2,π2 Let, u=2sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dudx=211-x2dudx=21-x2             ...iiiAnd, from equation ii,v=θ                                 Since, cos-1cosθ=θ, if θ0,πv=sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivdividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2


ii Let,    u=sin-12x1-x2     Put    x=sinθ          u=sin-12sinθ1-sin2θ          u=sin-12 sinθ cosθ            u=sin-1sin2θ                    ...iAnd,   Let,      v=sec-111-x2            v=sec-111-sin2θ               v=sec-11cosθ              v=sec-1secθ               v=cos-111cosθ                       Since, sec-1x=cos-11x             v=cos-1cosθ                ...iiHere,           x12,1   sinθ12,1          θπ4,π2So, from equation i,    u=2θ                                 Since, sin-1sinθ=θ, if θ-π2,π2 Let, u=2sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dudx=211-x2dudx=21-x2            ...iiiAnd, from equation ii,v=θ                                 Since, cos-1cosθ=θ, if θ0,πv=sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivdividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2

Page No 10.112:

Question 8:

Differentiate cos xsin x with respect to sin xcos x.

Answer:

Let,     u=cosxsinx

Taking log on both sides,

log u=logcosxsinxlog u=sinx logcosx

Differentiating it with respect to x using chain rule,

1ududx=sinxddxlog cosx+log cosxddxsinx             using product rule1ududx=sinx1cosxddxcosx+log cosxcosxdudx=utanx×-sinx+log logxcosxdudx=cosxsinxcosx logcosx-sinx tanx                             ...iLet,   v=sinxcosx

Taking log on both sides,

log v=logsinxcosxlog v=cosx logsinx

Differentiating it with respect to x using chain rule,

1vdvdx=cosxddxlogsinx+logsinxddxcosx            using product rule1vdvdx=cosx1sinxddxsinx+logsinx-sinxdvdx=vcotxcosx-sinx logsinxdvdx=sinxcosxcotxcosx-sinx logsinxdividing equationi by ii,dudv=cosxsinxcosx logcosx-sinx tanxsinxcosxcotxcosx-sinx logsinx

Page No 10.112:

Question 9:

Differentiate sin-1 2x1+x2 with respect to cos-1 1-x21+x2, if 0<x<1

Answer:

Let,   u=sin-12x1+x2Put   x=tanθ    u=sin-12tanθ1+tan2θ    u=sin-1sin2θ               ...iLet   v=cos-11-x21+x2    v=cos-11-tan2θ1+tan2θ    v=cos-1cos2θ          ...iiHere, 0<x<1     0<tanθ<1     0<θ<π4So, from equation i,u=2θ                            Since, sin-1sinθ=θ , if θ-π2,π2u=2tan-1x                  Since , x=tanθ

Differentiating it with respect to x,

dudx=21+x2                ...iiifrom equation ii,v=2θ                         Since, cos-1cosθ=θ, if θ0,πv=2tan-1x               Since, x=tanθ

Differentiating it with respect to x,

dvdx=21+x2                ...ivDividing equation iii by iv,dudxdvdx=21+x2×1+x22dudv=1



Page No 10.113:

Question 10:

Differentiate tan-1 1+ax1-ax with respect to 1+a2 x2

Answer:

Let,   u=tan-11+ax1-axPut   ax=tanθ    u=tan-11+tanθ1-tanθ    u=tan-1tanπ4+tanθ1-tanπ4tanθ    u=tan-1tanπ4+θ    u=π4+θ    u=π4+tan-1ax                  Since, tanθ=ax               

Differentiating it with respect to x,

dudx=0+11+ax2ddxax  dudx=a1+a2x2                  ...i              Now,Let,   v=1+a2x2

Differentiating it with respect to x,

dvdx=121+a2x2ddx1+a2x2dvdx=121+a2x22a2xdvdx=a2x1+a2x2                ...ii            Dividing equation i by ii,dudxdvdx=a1+a2x2×1+a2x2a2xdudv=1ax1+a2x2

Page No 10.113:

Question 11:

Differentiate sin-1 2x 1-x2 with respect to tan-1 x1-x2, if -12<x<12

Answer:

Let,   u=sin-12x1-x2Put   x=sinθ    u=sin-12sinθ1-sin2θ    u=sin-12 sinθcosθ     u=sin-1sin2θ                 ...iLet  v=tan-1x1-x2   v=tan-1sinθ1-sin2θ     v=tan-1sinθcosθ   v=tan-1tanθ                  ...iiHere,    -12<x<12       -12<sinθ<12       -π4<θ<π4So, from equation i,u=2θ                      Since, sin-1sinθ=θ, if θ-π2,π2u=2sin-1x             Since, x=sinθ

Differentiating it with respect to x,

dudx=21-x2            ...iiifrom equation ii,v=θ                           Since, tan-1tanθ=θ, if θ-π2,π2v=sin-1x                  Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivDividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2

Page No 10.113:

Question 12:

Differentiate tan-1 2x1-x2 with respect to cos-1 1-x21+x2, if 0<x<1

Answer:

Let,     u=tan-12x1-x2Put     x=tanθ      u=tan-12tanθ1-tan2θ      u=tan-1tan2θ               ...ilet,      v=cos-11-x21+x2     v=cos-11-tan2θ1+tan2θ      v=cos-1cos2θ                ... iiHere,  0<x<1       0<tanθ<1       0<θ<π4So, from equation i,u=2θ                          Since, tan-1tanθ=θ, if θ-π2,π2u=2tan-1x                 Since, x=tanθ


differentiating it with respect to x,

dudx=21+x2                 ...iiiFrom equation ii,v=θ                        Since, cos-1cosθ=θ, if θ0,πv=2tan-1x            Since, x=tanθ


Differentiating it with respect to x,


dvdx=21+x2                ...ivDividing equation iii by iv,dudxdvdx=21+x2×1+x22dudv=1

Page No 10.113:

Question 13:

Differentiate tan-1 x-1x+1 with respect to sin-1 3x-4x3, if -12<x<12

Answer:

Let,     u=tan-1x-1x+1Put     x=tanθ      u=tan-1tanθ-1tanθ+1      u=tan-1tanθ-tanπ41+tanθ tanπ4          u=tan-1tanθ-π4               ...i    Here,  -12<x<12       -12<tanθ<12       -tan-112<θ<tan-112So, from equation i,u=θ-π4                          Since, tan-1tanθ=θ, if θ-π2,π2u=tan-1x -π4                Since, x=tanθ

differentiating it with respect to x,

dudx=11+x2-0  dudx=11+x2               ...ii              And,Let,     v=sin-13x-4x3Put     x=sinθ      v=sin-13sinθ-4sin3θ      v=sin-1sin3θ          ...iiiNow,  -12<x<12    -12<sinθ<12    -16<θ<π6So, from equation iii,v=3θ                           Since, sin-1sinθ=θ, if θ-π2,π2v=3sin-1x                  Since, x=sinθ

Differentiating it with respect to x,

dvdx=31-x2                ...ivDividing equation iii by iv,dudxdvdx=11+x2×1-x23dudv=1-x231+x2

Page No 10.113:

Question 14:

Differentiate tan-1 cos x1+sin x with  respect to sec-1 x

Answer:

Let,    u=tan-1cosx1+sinx    u=tan-1tanπ4-x2    u=π4-x2

Differentiating it with respect to x,

dudx=0-12dudx=-12                ...iLet,    v=sec-1x

Differentiating it with respect to x,

dvdx=1xx2-1           ...iiDividing equation i by ii,dudxdvdx=-12×xx2-11dudv=-xx2-12

Page No 10.113:

Question 15:

Differentiate sin-1 2x1+x2 with respect to tan-1 2 x1-x2, if -1<x<1

Answer:

Let,    u=sin-12x1+x2Put     x=tanθ θ=tan-1x,      u=sin-12tanθ1+tan2θ      u=sin-1sin2θ                 ...iLet,    v=tan-12x1-x2      v=tan-12tanθ1-tan2θ      v=tan-1tan2θ                ...iiHere,  -1<x<1       -1<tanθ<1       -π4<tanθ<π4So, from equation i,u=2θ                          Since, sin-1sinθ=θ, if θ-π2,π2u=2tan-1x                


Differentiating it with respect to x,

dudx=21+x2             ...iiifrom equation ii,v=2θ                    Since, tan-1tanθ=θ , if θ-π2,π2v=2tan-1x

Differentiating it with respect to x,

dvdx=21+x2                 ...ivDividing equation iii by iv,dudxdvdx=21+x2       ×1+x22dudv=1

Page No 10.113:

Question 16:

Differentiate cos-1 4x3-3x with respect to tan-1 1-x2x, if 12<x<1

Answer:

Let,    u=cos-14x3-3xPut,    x=cosθ      θ=cos-1xNow,   u=cos-14cos3θ-3cosθ       u=cos-1cos3θ                     ...i Let,     v=tan-11-x2x      v=tan-11-cos2θcosθ        v=tan-1sinθcosθ      v=tan-1tanθ                          ...iiHere,        12<x<1   12<cosθ<1   0<θ<π3So, from equation i,u=3θ                              Since, cos-1cosθ=θ, if θ0,πu=3cos-1x

Differentiating it with respect to x,

dudx=-31-x2             ...iiiFrom equation ii,v=θ                     Since, tan-1tanθ=θ, if θ-π2,π2v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2               ...ivDividing equation iii by iv,dudxdvdx=-31-x2-1-x21dudv=3

Page No 10.113:

Question 17:

Differentiate tan-1 x1-x2 with respect to sin-1 2x 1-x2, if -12<x<12

Answer:

 Let,     u=tan-1x1-x2Put    x=sinθ      θ=sin-1x      u=tan-1sinθ1-sin2θ        u=tan-1sinθcosθ      u=tan-1tanθ                 ...iAndLet,    v=sin-12x1-x2           v=sin-12sinθ1-sin2θ           v=sin-12 sinθcosθ           v=sin-1sin2θ                 ...iiHere,        -12<x<12   -12<sinθ<12   -π4<θ<π4So, from equation i,u=θ                              Since, tan-1tanθ=θ, if θ-π2,π2u=sin-1x

Differentiating it with respect to x,

dudx=11-x2             ...iiifrom equation ii,v=2θ                     Since, sin-1sinθ=θ, if θ-π2,π2 v=2sin-1x 

Differentiating it with respect to x,

dvdx=21-x2               ...ivDividing equation iii by iv,dudxdvdx=11-x21-x22dudv=12

Page No 10.113:

Question 18:

Differentiate sin-1 1-x2 with respect to cot-1 x1-x2, if 0<x<1

Answer:

Let,    u=sin-11-x2Put    x=cosθ      θ=cos-1xWe get,     u=sin-1sinθ                     ...i Let,     v=cot-1x1-x2      v=cot-1cosθ1-cos2θ        v=cot-1cosθsinθ      v=cot-1cotθ                       ...iiHere,        0<x<1   0<cosθ<1   0<θ<π2So, from equation i,u=θ                              Since, sin-1sinθ=θ, ifθ-π2,π2u=cos-1x

Differentiating it with respect to x,

dudx=-11-x2             ...iiiFrom equation ii,v=θ                     Since, cot-1cotθ=θ, if θ0,πv=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2               ...ivDividing equation iii by iv,dudxdvdx=-11-x21-x2-1dudv=1

Page No 10.113:

Question 19:

Differentiate sin-1 2 ax 1-a2 x2 with respect to 1-a2 x2, if-12<ax<12

Answer:

 Let,     u=sin-12ax1-a2x2Put    ax=sinθ       θ=sin-1ax     u=sin-12sinθ1-sin2θ       u=sin-12 sinθcosθ     u=sin-1sin2θ                         ...iAndLet,           v=1-a2x2Differentiating it with respect to x,dvdx=121-a2x2 ×ddx1-a2x2       dvdx=0-2a2x21-a2x2    dvdx=-a2x1-a2x2                            ...iiHere,        -12<ax<12   -12<sinθ<12   -π4<θ<π4So, from equation i,u=2θ                              Since, sin-1sinθ=θ, if θ-π2,π2u=2sin-1x

Differentiating it with respect to x,

dudx=2×11-ax2ddxax  dudx=21-a2x2a  dudx=2a1-a2x2       ...iii      

Dividing equation iii by ii,dudxdvdx=2a1-a2x21-a2x2-a2xdudv=-2ax

Page No 10.113:

Question 20:

Differentiate tan-1 1-x1+x with respect to 1-x2, if-1<x<1

Answer:

Let,   u=tan-11-x1+xPut    x=tanθ     θ=tan-1x    u=tan-11-tanθ1+tanθ    u=tan-1tanπ4-θ               ...iHere,             -1<x<1     -1<tanθ<1     -π4<θ<π4     π4>-θ>π4     -π4<-θ<π4     0<π4-θ<π2So, from equation i,u=π4-θ                               Since, tan-1tanθ=θ, if θ-π2,π2u=π4-tan-1x

Differentiating it with respect to x,

dudx=0-11+x2dudx=-11+x2           ...iiAnd  let,      v=1-x2

Differentiating it with respect to x,

dvdx=121-x2×ddx1-x2dvdx=121-x2-2xdvdx=-x1-x2                     ...iiiDividing equation ii by iii,dudxdvdx=-11+x2×1-x2-xdudv=1-x2x1+x2



Page No 10.117:

Question 1:

If f (x) = logx2 (log x), the f' (x) at x = e is
(a) 0
(b) 1
(c) 1/e
(d) 1/2 e

Answer:

(d) 1/2 e

We have, fx=logx2logxfx=loglogxlogx2       fx=loglogx2 logxf'x=12×ddxloglogx logxf'x=12×1logx×1x×logx-loglogxx logx2f'x=12×1x-loglogxx logx2f'e=12×1e-loglogee loge2       Putting x=ef'e=12×1e1f'e=12e

Page No 10.117:

Question 2:

The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is
(a) xlog x

(b) log xx

(c) x log x-1

(d) none of these

Answer:

(c) x log x-1

We have,
fx=log xflogx=loglogxf'logx=1logxddxlogxf'logx=1x logxf'logx=x logx-1

Page No 10.117:

Question 3:

The derivative of the function cot-1 cos 2 x1/2 at x=π/6 is
(a) (2/3)1/2
(b) (1/3)1/2
(c) 31/2
(d) 61/2

Answer:

(a) (2/3)1/2
We have,  y=cot-1cos 2x

dydx=-11+cos 2xddxcos 2xdydx=-12 cos2x×12cos 2xddxcos 2xdydx=-12 cos2x×12cos 2x×-2sin 2xdydx=sin2xcos2x × 2cos2xdydx=2 sinx cosxcos2x × 2cos2xdydx=tanxcos2xSo, at x=π6, we getdydxx=π6=tanπ6cos2π6=1312=2312

Page No 10.117:

Question 4:

Differential coefficient of sec sec tan-1 x is
(a) x1+x2

(b) x 1+x2

(c) 11+x2

(d) x1+x2

Answer:

(d) x1+x2

We have, y=sectan-1xdydx=sectan-1x tantan-1x×ddxtan-1xdydx=sectan-1x tantan-1x×11+x2dydx=yx1+x2dydx=x1+x2 y
This is the equation of differential equation which have coefficient x1+x2.

Page No 10.117:

Question 5:

If fx=tan-1 1+sin x1-sin x, 0xπ/2, then f' π/6 is
(a) − 1/4
(b) − 1/2
(c) 1/4
(d) 1/2

Answer:

(d) 1/2

Let  y=tan-11+sinx1-sinx y=tan-11-cosπ2+x1+cosπ2+x y=tan-12 sin2π4+x22 cos2π4+x2  y=tan-1tanπ4+x2=π4+x2dydx=12

Page No 10.117:

Question 6:

If y=1+1xx, then dydx=

(a) 1+1xx 1+1x-1x+1

(b) 1+1xx log 1+1x

(c) x+1xx log x+1-xx+1

(d) x+1xx log 1+1x+1x+1

Answer:

(a) 1+1xx 1+1x-1x+1

Let y=1+1xxTaking log on both sides,log y=x log1+1x1ydydx=xddxlog1+1x+log1+1xddxx            1ydydx=x11+1xddx1+1x+log1+1x1ydydx=x × xx+1-1x2+log1+1x1ydydx=x2x+1×-1x2+log1+1x1ydydx=-1x+1+log1+1xdydx=y-1x+1+log1+1xdydx=1+1xxlog1+1x-1x+1



Page No 10.118:

Question 7:

If xy=ex-y, then dydx is
(a) 1+x1+log x

(b) 1-log x1+log x

(c) not defined

(d) log x1+log x2

Answer:

(d) log x1+log x2

We have, xy=ex-yTaking log on both sides we get, y logx=x-yloge ey logx=x-yy logx+y=xy1+logx=xy=x1+logx
dydx=1+logx × 1-x ×0+1x1+logx2dydx=1+logx -11+logx2dydx=logx1+logx2

Page No 10.118:

Question 8:

Given fx=4x8, then

(a) f'12=f'-12

(b) f12=-f'-12

(c) f-12=f-12

(d) f12=f'-12

Answer:

 c  f-12=f-12We have, fx=4x8f'x=32x7Now,  f12=4128=41256=164 f-12=4-128=41256=164f'12=32127=321128=14f'-12=32-127=-321128=-14

Page No 10.118:

Question 9:

If x=a cos3 θ, y=a sin3 θ, then 1+dydx2=
(a) tan2 θ
(b) sec2 θ
(c) sec θ
(d) sec θ

Answer:

(d) sec θ

We have, x=a cos3θdxdθ=addθcos3θdxdθ=3acos2θddθcosθdxdθ=-3acos2θsinθ                  ...1and,  y=a sin3θdydθ=addθsin3θdydθ=3a sin2θddθsinθdydθ=3a sin2θ cosθ                     ...2Dividing 2 by 1, we get, dydθdxdθ=3a sin2θ cosθ   -3acos2θsinθdydx=sinθ-cosθdydx=-tanθNow, 1+dydx2=1+tan2θ=sec2θ=secθ

Page No 10.118:

Question 10:

If y=sin-1 1-x21+x2, then dydx=

(a) -21+x2

(b) 21+x2

(c) 12-x2

(d) 22-x2

Answer:

(a) -21+x2

Let y=sin-11-x21+x2Differentiating with respect to x using chain rule, we get,dydx=11-1-x21+x22ddx1-x21+x2dydx=1+x21+x22-1-x221+x2ddx1-x2-1-x2ddx1+x21+x22               using quotient ruledydx=1+x21+x22-1-x221+x2-2x-1-x22x1+x22dydx=1+x22x-2x-2x3-2x+2x31+x22dydx=-4x2x1+x2dydx=-21+x2

Page No 10.118:

Question 11:

The derivative of sec-1 12 x2+1 w.r.t. 1+3 x at x=-1/3

(a) does not exist
(b) 0
(c) 1/2
(d) 1/3

Answer:

(a) does not exist

We know that sec-1α is not defined for α-1, 1Here for x=-13, 12x2+1=911-1, 1 sec-112x2+1 is not defined at  x=-13 Derivative of sec-112x2+1 does not exist at x=-13

Page No 10.118:

Question 12:

For the curve x+y=1, dydx at 1/4, 1/4 is
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer:

(c) −1

We have, x+y=1Differentiating with respect to x, we get,12x+12ydydx=012ydydx=-12xdydx=-12x×2y1dydx=-yxNow, dydx14,14=-1414=-1

Page No 10.118:

Question 13:

If sin x+y=log x+y, then dydx=
(a) 2
(b) − 2
(c) 1
(d) − 1]

Answer:

(d) − 1

We have, sinx+y=logx+ycosx+y1+dydx=1x+y1+dydxcosx+y+cosx+ydydx=1x+y+1x+ydydxcosx+ydydx-1x+ydydx=1x+y-cosx+ycosx+y-1x+ydydx=1x+y-cosx+y-1x+y-cosx+ydydx=1x+y-cosx+ydydx=-1

Page No 10.118:

Question 14:

Let =sin-1 2 x1+x2 and V=tan-1 2 x1-x2, then ddV=
(a) 1/2
(b) x
(c) 1-x2x2-4
(d) 1

Answer:

(d) 1
We have, u=sin-12x1+x2 and v=tan-12x1-x2dudx=21+x2  and dvdx=21+x2   

dudv=dudxdvdx=21+x2×1+x22=1

Page No 10.118:

Question 15:

ddx tan-1 cos x1+sin x equals

(a) 1/2
(b) − 1/2
(c) 1
(d) − 1

Answer:

(b) − 1/2

Let  u=tan-1cosx1+sinx   u=tan-1cos2x2-sin2x2cos2x2+sin2x2+2sinx2cosx2   u=tan-1cosx2-sinx2cosx2+sinx2cosx2+sinx22   u=tan-1cosx2-sinx2cosx2+sinx2  u=tan-11-tanx21+tanx2  u=tan-1tanπ4-tanx21+tanπ4×tanx2  u=tan-1tanπ4-x2  u=π4-x2

dudx=0-12dudx= -12



Page No 10.119:

Question 16:

ddx log ex x-2x+23/4 equals

(a) x2-1x2-4

(b) 1

(c) x2+1x2-4

(d) exx2-1x2-4

Answer:

(a) x2-1x2-4

Let y=ddxlogexx-2x+234 y=ddxxloge+34logx-2x+2 y=ddxx+34logx-2x+2dydx=1+34x-2x+2×x+2×1-x-2×1x+22dydx=1+3x+24x-2×x+2-x+2x+22dydx=1+3x+24x-2×4x+2dydx=1+3x2-4dydx=x2-4+3x2-4dydx=x2-1x2-4

Page No 10.119:

Question 17:

If y=sin x+y, then dydx=

(a) sin x2 y-1

(b) sin x1-2 y

(c) cos x1-2 y

(d) cos x2 y-1

Answer:

(d) cos x2 y-1

We have, y=sinx+ySquaring both sides, we get,y2=sinx+yy2-y=sinx2ydydx-dydx=cosxdydx2y-1=cosxdydx=cosx2y-1

Page No 10.119:

Question 18:

If 3 sin xy+4 cos xy=5, then dydx=

(a) -yx

(b) 3 sin xy+4 cos xy3 cos xy-4 sin xy

(c) 3 cos xy+4 sin xy4 cos xy-3 sin xy

(d) none of these

Answer:

(a) -yx

We have, 3 sinxy+4 cosxy=5     3 cosxyxdydx+y-4 sinxyxdydx+y=0   xdydx+y3 cosxy-4 sinxy =0   xdydx+y=0   xdydx=-y     dydx=-yx

Page No 10.119:

Question 19:

If sin y=x sin a+y, then dydx is

(a) sin asin a sin2 a+y

(b) sin2 a+ysin a

(c) sin a sin2 a+y

(d) sin2 a-ysin a

Answer:

(b) sin2 a+ysin a

We have, sin y=x sina+y

ddxsin y=ddxx sina+ycos ydydx=sina+yddxx+xddxsina+ycos ydydx=sina+y×1+x cosa+ydydx cos ydydx=sina+y+x cosa+ydydxcos ydydx-x cosa+ydydx=sina+ycos y-x cos a+ydydx=sina+ycos y-sin ysina+y×cosa+ydydx=sina+y                       sin y=2 sinx cosxx=sin ysina+ysina+y cos y-sin y cosa+ysina+ydydx=sina+ysina+y-ysina+y×dydx=sina+y dydx=sin2a+ysina   

Page No 10.119:

Question 20:

The derivative of cos-1 2x2-1 with respect to cos-1 x is
(a) 2

(b) 12 1-x2

(c) 2/x

(d) 1-x2

Answer:

(a) 2

Let u=cos-12x2-1Put x=cosθθ=cos-1xdθdx=-11-x2Now, u=cos-1cos2θu=2θ

dudx=2dθdxdudx=-21-x2               ...iand, v=cos-1xv=cos-1cosθv=θ

dvdx=dθdxdvdx=-11-x2                       ...iiDividing i by ii, we get,dudxdvdx=-21-x2×1-x2-1dudv=2

Page No 10.119:

Question 21:

If fx=x2+6x+9, then f'x is equal to
(a) 1 for x<-3
(b) -1 for x<-3
(c) 1 for all xR
(d) none of these

Answer:

b-1  for x<-3 We have, fx=x2+6x+9                        = x+32                           =x+3 fx=x+3,       x-3 -x-3,      x<-3   f'x=1,       x-3 -1,      x<-3   f'x=-1  for x<-3  

Page No 10.119:

Question 22:

If fx=x2-9x+20, then f' (x) is equal to
(a) -2x+9 for all xR
(b) 2x-9 if 4<x<5
(c) -2x+9, if 4<x<5
(d) none of these

Answer:

c-2x+9 for 4<x<5 We have, fx=x2-9x+20 fx=x2-9x+20,       -<x4-x2-9x+20,       4<x<5x2-9x+20,       5x<fx=2x-9,       -<x4-2x+9,       4<x<52x-9,       5x<f'x=-2x+9 for 4<x<5    

Page No 10.119:

Question 23:

If fx=x2-10x+25, then the derivative of f (x) in the interval [0, 7] is
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(d) none of these

We have, fx=x2-10x+25=x-52=x-5 =x-5  for x>5 -x-5 for x<5LHD=limx5- fx-fax-a        =limx5-x2-10x+25-52-105+25x-5        =limx5-x-5x-5        =limx5--x-5x-5        =-1RHD=limx5+ fx-fax-a        =limx5+x2-10x+25-52-105+25x-5        =limx5+x-5x-5        =limx5+x-5x-5        =1Here, LHDRHDThus, the functionis not differentiable at x=5

Page No 10.119:

Question 24:

If fx=x-3 and gx=fof x, then for x > 10, g ' (x) is equal to
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(c) 0

For, x>10fx=x-3=x-3gx=fof x=x-3-3                       =x-3-3                       =x-6g'x =1

Page No 10.119:

Question 25:

If fx=xlxml+m xmxnm+n xnxln+1, the f' (x) is equal to
(a) 1
(b) 0
(c) xl+m+n
(d) none of these

Answer:

(b) 0
We have, fx=xlxml+m xmxnm+n xnxln+1

fx=xl-ml+m × xm-nm+n × xn-ln+lfx=xl2-m2 × xm2-n2 × xn2-l2fx=xl2-m2+m2-n2+n2-l2fx=x0fx=1f'x=0    

Page No 10.119:

Question 26:

If y=11+xa-b+c-b+11+xb-c+xa-c+11+xb-a+xc-a, then dydx is equal to
(a) 1
(b) a+b+cxa+b+c-1
(c) 0
(d) none of these

Answer:

(c) 0


We have,y=11+xa-b+xc-b+11+xb-c+xa-c+11+xb-a+xc-a=11+xaxb+xcxb+11+xbxc+xaxc+11+xbxa+xcxa=xbxb+xa+xc+xcxc+xb+xa+xaxa+xb+xc=xbxa+xb+xc+xcxa+xb+xc+xaxa+xb+xc=xb+xc+xaxa+xb+xc=xa+xb+xcxa+xb+xc=1dydx=ddx1=0



Page No 10.120:

Question 27:

If 1-x6+1-y6=a3 x3-y3, then dydx is equal to
(a) x2y2 1-y61-x6

(b) y2x21-y61+x6

(c) x2y21-x61-y6

(d) none of these

Answer:

(a) x2y2 1-y61-x6

We have,   1-x6+1-y6=ax3-y3Putting x3=sinA and y3=sinB1-sin2A+1-sin2B=asinA-sinB cosA+cosB=asinA-sinB2cosA+B2cosA-B2=2a sinA-B2cosA+B2cotA-B2=aA-B2=cot-1a A-B=2 cot-1asin-1x3-sin-1y3=2 cot-1a

11-x6×ddxx3-11-y6×ddxy3=011-x6×3x2-11-y6×3y2×dydx=0dydx=x2y21-y61-x6

Page No 10.120:

Question 28:

If y=log tan x, then the value of dydxat x=π4 is given by
(a) ∞
(b) 1
(c) 0
(d) 12

Answer:

(b) 1

We have,  y=logtanxdydx=1tanx×ddxtanxdydx=1tanx×12tanx×ddxtanxdydx=sec2x2 tanxNow, dydxx=π4=secπ422 tanπ4=22×1=1

Page No 10.120:

Question 29:

If sin-1 x2-y2x2+y2=log a then dydx is equal to

(a) x2-y2x2+y2

(b) yx

(c) xy

(d) none of these

Answer:

(b) yx
We have, sin-1x2-y2x2+y2=loga  x2-y2x2+y2=sin loga

x2+y22x-2ydydx-x2-y22x+2ydydxx2+y22=02x3-2x2ydydx+2xy2-2y3dydx-2x3-2x2ydydx+2xy2+2y3dydxx2+y22=0-4x2ydydx+4xy2=0-4x2ydydx=-4xy2dydx=4xy24x2ydydx=yx

Page No 10.120:

Question 30:

If sin y=x cos a+y, then dydx is equal to
(a) cos2 a+ycos a

(b) cos acos2 a+y

(c) sin2 ycos a

(d) none of these

Answer:

(a) cos2 a+ycos a

We have, sin y=x cosa+y

ddxsin y=ddxx cosa+ycos ydydx=1×cosa+y-x sina+yddxa+ycos ydydx=cosa+y-x sina+ydydxcos ydydx+x sina+ydydx=cosa+ycos y+x sina+ydydx=cosa+ycos y+sin ycosa+y×sina+ydydx=cosa+y                         sin y=x cosa+yx=sin ycosa+ycosa+y cos y+sin y sina+ycosa+ydydx=cosa+ycosa+y-ycosa+y×dydx=cosa+ydydx=cos2a+ycos a

Page No 10.120:

Question 31:

If y=log 1-x21+x2, then dydx=

(a) 4x31-x4

(b) -4x1-x4

(c) 14-x4

(d) -4x31-x4

Answer:

(b) -4x1-x4
We have, y=log1-x21+x2dydx=11-x21+x2ddx1-x21+x2dydx=1+x21-x21+x2-2x-1-x22x1+x22dydx=11-x2-2x-2x3-2x+2x31+x2dydx=-4x1-x4

Page No 10.120:

Question 32:

If y=tan-1 sin x+cos xcos x-sin x, then dydx is equal to
(a) 12
(b) 0
(c) 1
(d) none of these

Answer:

(c) 1

We have, y=tan-1sinx+cosxcosx-sinxdydx=11+sinx+cosxcosx-sinx2ddxsinx+cosxcosx-sinxdydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxddxsinx+cosx-sinx+cosxddxcosx-sinxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxcosx-sinx-sinx+cosx-sinx-cosxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxcosx-sinx+sinx+cosxsinx+cosxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2×cosx-sinx2+sinx+cosx2cosx-sinx2dydx=1

Page No 10.120:

Question 1:

If y = x x, then dydxx=-1=______________________.

Answer:


We know

x=x,x0-x,x<0

y=xx=x2,x0-x2,x<0

dydx=2x,x0-2x,x<0

For x < 0,

dydx=-2x

dydxx=-1=-2×-1=2


If y = xx, then dydxx=-1=       2      .

Page No 10.120:

Question 2:

If y = 2xx, then dydxx=-1=_________________ and dydxx=1=____________________.

Answer:


We know

x=x,x0-x,x<0

y=2x+x=2x+x,x02x-x,x<0

y=2x+x=3x,x0x,x<0

For x ≥ 0,

y = 3x

dydx=3

dydxx=1=3

For x < 0,

y = x

dydx=1

dydxx=-1=1

Thus, if y = 2x + |x|, then dydxx=-1=1 and dydxx=1=3.


If y = 2xx, then dydxx=-1=     1      and dydxx=1=     3     .

Page No 10.120:

Question 3:

If f(x) = x2-x, then f'(2)=_________________.

Answer:


fx=x2-x=x2-x,x2-x0-x2-x,x2-x<0

fx=x2-x=x2-x,xx-10x-x2,xx-1<0

fx=x2-x=x2-x,-<x0 or 1x<x-x2,0<x<1

For x ≥ 1,

fx=x2-x

f'x=2x-1

f'2=2×2-1=3



If f(x) = x2-x, then f'(2)=      3      .



Page No 10.121:

Question 4:

If y = sinxo and dydx = k cos xo , then k = ________________.

Answer:


y=sinx°

y=sinπ180x                      180°=π radians

Differentiating both sides with respect to x, we get

dydx=ddxsinπ180x

dydx=cosπ180x×ddxπ180x

dydx=cosx°×π180

dydx=π180cosx°

Comparing with dydx=kcosx°, we get

k=π180


If y = sinxo and dydx = k cosxo , then k =       π180       .

Page No 10.121:

Question 5:

If f(x) = exg(x), g(0) = 2, g'(0) = 1, then f'(0) = __________________.

Answer:


fx=exgx

Differentiating both sides with respect to x, we get

ddxfx=ddxexgx

f'x=ex×ddxgx+gx×ddxex

f'x=ex×g'x+gx×ex

Putting x = 0, we get

f'0=e0×g'0+g0×e0

f'0=1×1+2×1                 [g(0) = 2, g'(0) = 1 and e0 = 1]

f'0=1+2=3

Thus, the value of f'0 is 3.


If f(x) = exg(x), g(0) = 2, g'(0) = 1, then f'(0) = ____3____.

Page No 10.121:

Question 6:

If â€‹f(x) = 3x+2, then '(-3) = ___________________.

Answer:


We know

x+2=x+2,x-2-x+2,x<-2

fx=3x+2=3x+2,x-2-3x+2,x<-2

f'x=3,x-2-3,x<-2

For x < −2, f'x=-3

f'-3=-3



If â€‹f(x) = 3x+2, then '(−3) = ____−3____.

Page No 10.121:

Question 7:

If f(1) = 3, f'(2) = 1, then ddxIn fex+2x= ____________________________.

Answer:


Disclaimer: The solution is provided for the following question.

If f(1) = 3, f'(1) = 1, then ddxIn fex+2x= ____________________________.


Solution:

ddxlnfex+2x=1fex+2x×ddxfex+2x=1fex+2x×f'ex+2xddxex+2x=1fex+2x×f'ex+2x×ex+2

Putting x = 0, we get

ddxlnfex+2xx=0

=1fe0+2×0×f'e0+2×0×e0+2

=1f1×f'1×3            e0=1

=13×1×3                         f1=3, f'1=1

= 1


If f(1) = 3, f'(1) = 1, then ddxIn fex+2x=      1      .

Page No 10.121:

Question 8:

If f(x) = x x, then â€‹f '(x) = _________________.

Answer:


We know

x=x,x0-x,x<0

fx=xx=x2,x0-x2,x<0

f'x=2x,x0-2x,x<0

Thus, f'x=2x when x ≥ 0 and f'x=-2x when x < 0.


If f(x) = xx, then â€‹f '(x) = 2x when x ≥ 0 and −2x when x < 0.

Page No 10.121:

Question 9:

​If f(x) = x-1+x-3, then f '(2) = ______________________.

Answer:


We have

x-1=x-1,x1-x-1,x<1

x-3=x-3,x3-x-3,x<3

fx=x-1+x-3=-x-1-x-3,x<1x-1-x-3,1x<3x-1+x-3,x3

fx=x-1+x-3=-2x+4,x<12,1x<32x-4,x3

For 1 ≤ x < 3, f(x)  = 2

f'x=0, for 1 ≤ x < 3

f'2=0

Thus, the value of f '(2) is 0.


​If f(x) = x-1+x-3, then f '(2) = ___0___.

Page No 10.121:

Question 10:

​If f(x) = cosx-sinx, then f'π3=___________________.

Answer:


For π4<xπ2,

cosx<sinx

cosx-sinx<0

fx=cosx-sinx=-cosx-sinx

fx=-cosx+sinx

Differentiating both sides with respect to x, we get

ddxfx=ddx-cosx+ddxsinx

f'x=--sinx+cosx

f'x=sinx+cosx

f'π3=sinπ3+cosπ3=32+12=3+12         π4<π3π2


​If f(x) = cosx-sinx, then f'π3=           3+12          .

Page No 10.121:

Question 11:

​If f(x) = cos x, then f'π4 =______________________.

Answer:


For 0x<π2,

cosx > 0

fx=cosx=cosx

f'x=-sinx

f'π4=-sinπ4=-12


​If f(x) = cos x, then f'π4 =           -12            .

Page No 10.121:

Question 12:

The derivative of x2 with respect to x3 is __________________.

Answer:


Let u(x) = x2 and v(x) = x3.

ux=x2

dudx=2x

vx=x3

dvdx=3x2

dudv=dudxdvdx

dudv=2x3x2=23x

Thus, the derivative of x2 with respect to x3 is 23x.


The derivative of x2 with respect to x3 is      23x     .












 

Page No 10.121:

Question 13:

For the curve x+y=1, dydx at 14,14 is ______________________.

Answer:


x+y=1         (Given)

Differentiating both sides with respect to x, we get

ddxx+ddxy=ddx1

12x+12ydydx=0

12ydydx=-12x

dydx=-yx=-yx

dydxπ4,π4=-1414

dydxπ4,π4=-1

Thus, the value of dydx at π4,π4 is −1.


For the curve x+y=1, dydx at 14,14 is      -1        .

Page No 10.121:

Question 14:

​If f(x) = sin x, then f'-π4 = ____________________.

Answer:


For -π2<x<0,

sinx<0

fx=sinx=-sinx

f'x=-cosx

f'-π4=-cos-π4=-cosπ4=-12                 cos-θ=cosθ


If f(x) = sin x, then f'-π4 =              -12              .

Page No 10.121:

Question 15:

​If f(x) = sin x - cosx, then f'π6 = ________________________.

Answer:


For 0<x<π4,

sinx<cosx

sinx-cosx<0

fx=sinx-cosx=-sinx-cosx

fx=-sinx+cosx

Differentiating both sides with respect to x, we get

f'x=-cosx-sinx

f'π6=-cosπ6-sinπ6=-32-12=-123+1


If f(x) = sin x - cosx, then f'π6 =                -123+1                 .

Page No 10.121:

Question 16:

If y = tan xo, then dydxx=45o = ________________________.

Answer:


y=tanx°

y=tanπ180x           180°=π radians

Differentiating both sides with respect to x, we get

dydx=ddxtanπ180x

dydx=sec2π180x×ddxπ180x

dydx=π180sec2π180x

dydx=π180sec2x°

dydxx=45°=π180sec245°

dydxx=45°=π180×22=π90


If y = tanxº, then dydxx=45o =      π90     .

Page No 10.121:

Question 17:

If y = sin-1(ex) + cos-1(ex), then dydx = ____________________.

Answer:


y=sin-1ex+cos-1ex

y=π2          sin-1x+cos-1x=π2

Differentiating both sides with respect to x, we get

dydx=ddxπ2

dydx=0


If y = sin−1(ex) + cos−1(ex), then dydx = ____0____.

Page No 10.121:

Question 18:

If y = sin-1(3x-4x3), 12< 1, then dydx = ______________________.

Answer:


y = sin−1(3x − 4x3)

Let x = sinθ.

y=sin-13sinθ-4sin3θ

y=sin-1sin3θ

Now,

12<x<1

12<sinθ<1

π6<θ<π2

π2<3θ<3π2

y=sin-1sin3θ

y=sin-1sinπ-3θ

y=π-3θ              π2<3θ<3π2-π2<π-3θ<π2

y=π-3sin-1x

Differentiating both sides with respect to x, we get

dydx=0-3×11-x2dydx=-31-x2


If y = sin−1(3x − 4x3), 12< 1, then dydx =       -31-x2      .

Page No 10.121:

Question 19:

If y = sec-1 x+1x-1+sin-1x-1x+1, then dydx is equal to ___________________.

Answer:


We know

sec-1a=cos-11a

sec-1x+1x-1=cos-1x-1x+1      .....(1)

Now,

y=sec-1x+1x-1+sin-1x-1x+1

y=cos-1x-1x+1+sin-1x-1x+1        [Using (1)]

y=π2            sin-1a+cos-1a=π2

Differentiating both sides with respect to x, we get

dydx=0

Thus, if y=sec-1x+1x-1+sin-1x-1x+1, then dydx=0.


If y=sec-1x+1x-1+sin-1x-1x+1 , then dydx is equal to ___0___.

Page No 10.121:

Question 20:

The derivative of cos x with respect to sin x is __________________.

Answer:


Let u(x) = cosx and v(x) = sinx.

ux=cosx

dudx=ddxcosx

dudx=-sinx     .....(1)

vx=sinx

dvdx=ddxsinx

dvdx=cosx        .....(2)

dudv=dudxdvdx

dudv=-sinxcosx       [From (1) and (2)]

dudv=-tanx

Thus, the derivative of cosx with respect to sinx is −tanx.


The derivative of cos x with respect to sin x is ___−tanx___.

Page No 10.121:

Question 21:

The derivative of log10x with respect to x is ___________________.

Answer:


Let y=log10x.

y=log10x

y=logexloge10            logba=logcalogcb

Differentiating both sides with respect to x, we get

dydx=ddxlogexloge10

dydx=1loge10×ddxlogex

dydx=1loge10×1x

dydx=1xloge10

Thus, the derivative of log10x with respect to x is 1xloge10.

The derivative of log10x with respect to x is       1xloge10      .

Page No 10.121:

Question 22:

Ifddx(f(x))=11+x2, then ddxfx3=_________________________.

Answer:


ddxfx3=f'x3×ddxx3

ddxfx3=f'x3×3x2     .....(1)

Now,

ddxfx=f'x=11+x2          (Given)

Replacing x by x3, we get

f'x3=11+x32

f'x3=11+x6         .....(2)

From (1) and (2), we get

ddxfx3=11+x6×3x2

ddxfx3=3x21+x6


Ifddx(f(x))=11+x2, then ddxfx3=     3x21+x6     .

Page No 10.121:

Question 23:

If y = cos (sin x2), then dydx at x=π2 is equal to ______________________.

Answer:


y=cossinx2

Differentiating both sides with respect to x, we get

dydx=ddxcossinx2

dydx=-sinsinx2×ddxsinx2

dydx=-sinsinx2×cosx2×ddxx2

dydx=-sinsinx2×cosx2×2x

dydx=-2xcosx2sinsinx2

Putting x=π2, we get

dydxx=π2=-2×π2×cosπ22sinsinπ22

dydxx=π2=-2×π2×cosπ2×sinsinπ2

dydxx=π2=-2×π2×0×sin1

dydxx=π2=0

Thus, dydx at x=π2 is 0.


If y = cos (sin x2), then dydx at x=π2 is equal to ___0___.

Page No 10.121:

Question 24:

If y = logx, x0, then dydx = ___________________.

Answer:


For y=logxto be defined,

x0

x0

Now,

y=logx=logx,x>0log-x,x<0

dydx=1x,x>01-x×-1,x<0

dydx=1x,x>01x,x<0

dydx=1x, x0


If y = logx, x0, then dydx =        1x       .

Page No 10.121:

Question 25:

If f(x) = ax2 + bx + c, then f '(1) + f '(4) - '(5) is equal to _____________________.

Answer:


fx=ax2+bx+c

f'x=ddxax2+bx+c

f'x=2ax+b

f'1+f'4-f'5

=2a×1+b+2a×4+b-2a×5+b

=2a+b+8a+b-10a-b

=b

Thus, the value of f'1+f'4-f'5 is b.

If f(x) = ax2 + bx + c, then f '(1) + f '(4) − '(5) is equal to ___b___.

Page No 10.121:

Question 26:

If '(1) = 2 and g'2 = 4, then the derivative of f(tan x) with respect of g(secx) at xπ4 is equal to ______________.

Answer:


Let u(x) = f(tanx) and v(x) = g(secx).

ux=ftanx

dudx=ddxftanx

dudx=f'tanxddxtanx

dudx=f'tanx×sec2x          .....(1)

vx=gsecx

dvdx=ddxgsecx

dvdx=g'secxddxsecx

dvdx=g'secx×secxtanx           .....(2)

dudv=dudxdvdx

dudv=sec2x×f'tanxsecxtanx×g'secx

dudv=secx×f'tanxtanx×g'secx

Putting x=π4, we get

dudvx=π4=secπ4×f'tanπ4tanπ4×g'secπ4

dudvx=π4=2×f'11×g'2

dudvx=π4=2×21×4

dudvx=π4=12

Thus, the derivative of f(tan x) with respect of g(secx) at x = π4 is 12.


If '(1) = 2 and g'2 = 4, then the derivative of f(tan x) with respect of g(secx) at x = π4 is equal to      12     .



Page No 10.122:

Question 1:

If f (x) = loge (logex), then write the value of f' (e).

Answer:

We have,  fx=loge logex
Differentiating with respect to x,
f'x=1logexddxlogex   f'x=1logex1xf'e=1logee1e               x=ef'e=1e                  loge e=1

Page No 10.122:

Question 2:

If fx=x+1, then write the value of ddx fof x.

Answer:

We have,  fx=x+1 Now, fofx=ffx  fofx=fx+1fofx=x+1+1fof=x+2

ddxfofx=ddxx+ddx2ddxfofx=1+0ddxfofx=1

Page No 10.122:

Question 3:

If f'1=2 and y=f loge x, find dydx at x=e.

Answer:

We have,  f'1=2 and y=flogex
Differentiate it with respect to x,
dydx=f'logex×ddxlogexdydx=f'logex1xdydx=f'logee1e         x=edydx=f'11e                  logee=1dydx=2e                            f'1=2

Page No 10.122:

Question 4:

If f1=4, f'1=2, find the value of the derivative of log fex w.r. to x at the point x = 0.

Answer:

We have, f1=4 and f'1=2Let y=logfex

dydx=ddxlogfexdydx=1fex×ddxfexdydx=1fex×f'ex×ddxexdydx=exf'exfexPutting x=0, we get,dydx=e0f'e0fe0dydx=1f'1f1dydx=24                             f'1=2 and f1=4dydx=12

Page No 10.122:

Question 5:

If f'x=2x2-1 and y=f x2, then find dydx at x=1.

Answer:

We have, f'x=2x2-1and  y=fx2

dydx=ddxfx2dydx=f'x2ddxx2dydx=f'x2× 2xdydx=2xf'x2Putting x=1, we get,dydx=21f'12dydx=2 ×f'1dydx=2 ×1                               f'1=212-1=2-1=1dydx=2

Page No 10.122:

Question 6:

Let g (x) be the inverse of an invertible function f (x) which is derivable at x = 3. If f (3) = 9 and f' (3) = 9, write the value of g' (9).

Answer:

We have, f3=9 ,f'3=9and gx=f-1xgofx=xgfx=x

ddxgfx=1g'fxddxfx=1g'fx × f'x=1Puting x=3, we get,g'f3 × f'3=1g'9 × 9=1                               f3=9 , f'3=9g'9=19

Page No 10.122:

Question 7:

If y=sin-1 sin x,-π2xπ2. Then, write the value of dydx for x  -π2, π2.

Answer:

We have, y=sin-1sinxy=x                          sin-1sinx=x , if x-π2,π2

dydx=ddxxdydx=1

Page No 10.122:

Question 8:

If π2x3π2 and y=sin-1 sin x, find dydx.

Answer:

We have, y=sin-1sinx                               y=π-x                        sin-1sinx=π-x , if xπ2,3π2                          

dydx=ddxπ-xdydx=0-1dydx=-1

Page No 10.122:

Question 9:

If πx2π and y=cos-1 cos x, find dydx.

Answer:

We have, y=cos-1cosx   y=2π-x                               cos-1cosx=2π-x , if xπ,2π                       

dydx=ddx2π-xdydx=0-1dydx=-1

Page No 10.122:

Question 10:

If y=sin-1 2x1+x2, write the value of dydxfor x>1.

Answer:

We have, y=sin-12x1+x2Putting x=tanθ1<tanθ<π4<θ<π2    π2<2θ<πy=sin-1sin2θ  y=sin-1sinπ-2θ   y=π-2θ  y=π-2tan-1xdydx=0-21+x2  dydx=-21+x2                        
 

Page No 10.122:

Question 11:

If f0=f1=0, f'1=2 and y=f ex ef x, write the value of dydx at x=0.

Answer:

We have, f0=f1=0 , f'1=2and,y=fexefx

dydx=ddxfex × efxdydx=fexddxefx+efxddxfex                   Using product ruledydx=fex × efxddxfx+efx×f'exddxexdydx=fex × efx×f'x+efx×f'ex×exPutting x=0, we get,dydx=fe0 × ef0×f'0+ef0×f'e0×e0dydx=f1ef0×f'0+ef0×f'1×1dydx=0×e0×f'0+e0×2×1                             fx=f1=0 and f'1=2dydx=0+1×2×1dydx=2

Page No 10.122:

Question 12:

If y=x x, find dydx for x<0.

Answer:

We have, y=xxy=x-x  x<0y=-x2

dydx=ddx-x2dydx=-2x

Page No 10.122:

Question 13:

If y=sin-1 x+cos-1 x, find dydx.

Answer:

We have, y=sin-1x+cos-1x y=π2                             sin-1x+cos-1x=π2         

dydx=0

Page No 10.122:

Question 14:

If x=a θ+sin θ, y=a 1+cos θ, find dydx.

Answer:

We have, x=aθ+sinθ   and y=a1+cosθdxdθ=addθθ+ddθsinθ and dydθ=a0-sinθdxdθ=a1+cosθ and dydθ=-asinθ dydx=dydθdxdθ=-asinθa1+cosθ=-2sinθ2cosθ22 cos2θ2=-tanθ2          
 

Page No 10.122:

Question 15:

If -π2<x<0 and y=tan-1 1-cos 2x1+cos 2x, find dydx.

Answer:

We have, y=tan-11-cos2x1+cos2xy=tan-12 sin2x2 cos2xy=tan-1tan2xy=tan-1tanx                         tan-1tanx=-x , if x-π2,0y=-x

dydx=-1

Page No 10.122:

Question 16:

If y=xx, find dydx at x=e.

Answer:

We have, y=xx     ...i
Taking log on both sides,
log y=log xxlog y=x logx

1ydydx=xddxlogx+logxddxx1ydydx=x1x+logx 11ydydx=1+logxdydx=y1+logxdydx=xx1+logx                                using equation iPuting x=e, we get,dydx=ee1+logeedydx=ee1+1                                      logee=1dydx=2ee 



Page No 10.123:

Question 17:

If y=tan-1 1-x1+x, find dydx.

Answer:

We have, y=tan-11-x1+x

dydx=11+1-x1+x2ddx1-x1+xdydx=1+x21+x2+2x+1+x2-2x1+xddx1-x-1-xddx1+x1+x2                  using quotient ruledydx=1+x22x2+21+x-1-1-x11+xdydx=1+x22x2+1-x-1-1+x1+x2dydx=1+x22x2+1×-21+x2dydx=-1x2+1

Page No 10.123:

Question 18:

If y=loga x, find dydx.

Answer:

We have, y=logaxy=logxloga                      logab=logbloga

dydx=1logaddxlogxdydx=1loga1xdydx=1x loga

Page No 10.123:

Question 19:

If y=log tan x, write dydx.

Answer:

We have, y=logtanxy=logtanx12y=12log tanx                      logab=bloga

dydx=12×1tanxddxtanxdydx=12×1tanxsec2xdydx=12sinxcosx×cos2xdydx=12 sinx cosxdydx=1sin2xdydx=cosec2x

Page No 10.123:

Question 20:

If y=sin-1 1-x21+x2+cos-1 1-x21+x2, find dydx.

Answer:

We have, y=sin-1 1-x21+x2+cos-1 1-x21+x2y=π2       sin-1x+cos-1x=π2   

dydx=0

Page No 10.123:

Question 21:

If y=sec-1 x+1x-1+sin-1 x-1x+1, then write the value of dydx.

Answer:

We have, y=sec-1x+1x-1+sin-1x-1x+1                    y=cos-1x-1x+1+sin-1x-1x+1                   sec-1x=cos-11xy=π2      sin-1x+cos-1x=π2

dydx=0

Page No 10.123:

Question 22:

If x <1 and y=1+x+x2+... to ∞, then find the value of dydx.

Answer:

We have, y=1+x+x2+....to  y=11-x                            It is a G.P with first term  1and common ratio x

dydx=ddx11-xdydx=-11-x2ddx1-xdydx=-11-x2-1dydx=11-x2

Page No 10.123:

Question 23:

If u=sin-1 2x1+x2 and v=tan-1 2x1-x2, where -1<x<1, then write the value of dudv.

Answer:

We have, u=sin-12x1+x2 and v=tan-12x1-x2dudx=21+x2 and dvdx=21+x2              dudv=dudxdvdx=21+x2×1+x22=1

Page No 10.123:

Question 24:

If fx=log u xv x, u 1=v 1 and u' 1=v' 1=2, then find the value of f' (1).

Answer:

We have, fx=loguxvxand, u1=v1 , u'1=v'1=2            ...i

f'x=ddxloguxvxf'x=1uxvx×ddxuxvxf'x=vxux×vxddxux-uxddxvxvx2                     f'x=vxux×vx×u'x-ux×v'xvx2Putting x=1, we get,f'1=v1u1×v1×u'1-u1×v'1v12f'1=1×u1×2-u1×2u12      Using eqn 1f'1=0u12             f'1=0

Page No 10.123:

Question 25:

If y=log 3x, x0, find dydx.

Answer:

We have, y=log3x
dydx=ddxlog3xdydx=13xddx3xdydx=13x3dydx=1x

Page No 10.123:

Question 26:

If f (x) is an even function, then write whether f' (x) is even or odd.

Answer:

We have, fx is an even function.f-x=fx

ddxf-x=ddxfxf'-xddx-x=f'xf'-x×-1=f'x-f'-x=f'xf'-x=-f'xThus, f'x is an odd function.

Page No 10.123:

Question 27:

If f (x) is an odd function, then write whether f' (x) is even or odd.

Answer:

We have, fx is an odd function.f-x=-fx

ddxf-x=-ddxfxf'-xddx-x=-f'xf'-x×-1=-f'x-f'-x=-f'xf'-x=f'xThus, f'x is an even function.

Page No 10.123:

Question 28:

If x=3sint-sin3t, y=3cost-cos3t finddydx at t=π3

Answer:

x=3sint-sin3t and y=3cost-cos3tdxdt=3cost-3cos3t and dydt=-3sint+3sin3tdydx=dydtdxdt=-3sint+3sin3t3cost-3cos3tdydxt=π3=-3sinπ3+3sinπ3cosπ3-3cosπ=-3×32+03×12+3=-33292=-13

Page No 10.123:

Question 29:

If y = log (cos ex), then find dydx.

Answer:


y=logcosex

Differentiating both sides with respect to x, we get

dydx=ddxlogcosex

dydx=1cosex×ddxcosex

dydx=1cosex×-sinex×ddxex

dydx=-sinexcosex×ex

dydx=-extanex

 

Page No 10.123:

Question 30:

If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find ddxfogx.

Answer:


The given functions are f(x) = x + 7 and g(x) = x – 7, x ∈ R.

fogx

=fgx

=gx+7                  fx=x+7

=x-7+7                 gx=x-7

=x

ddxfogx=ddxx

ddxfogx=1

Thus, the value of ddxfogx is 1.


 



Page No 10.17:

Question 1:

Differentiate the following functions from first principles:

ex

Answer:

Let  fx=e-xfx+h=e-x+h ddxfx=limh0fx+h-fxh                  =limh0e-x+h-e-xh                  =limh0e-x×e-h-e-xh                  =limh0e-xe-h-1-h×-1                  =-e-x limh0e-h-1-h                                                         =-e-x                    limh0e-h-1-h=1So, ddxe-x=-e-x

Page No 10.17:

Question 2:

Differentiate the following functions from first principles:

e3x

Answer:

Let  fx=e3xfx+h=e3x+hddxfx=limh0fx+h-fxh                =limh0e3x+h-e3xh                =limh0e3xe3h-e3xh                =limh0e3xe3h-13h×3                =3e3xlimh0e3h-13h                =3e3xHence, ddxe3x=3e3x

Page No 10.17:

Question 3:

Differentiate the following functions from first principles:

eax+b

Answer:

Let  fx=eax+b       fx+h=eax+h+b        ddxfx=limh0fx+h-fxh                             =limh0eax+h+b-eax+bh                             =limh0eax+beah-eax+bh                             =limh0eax+beah-1ah×a                             =aeax+blimh0eah-1ah                             =aeax+bSo, ddxeax+b=aeax+b

Page No 10.17:

Question 4:

Differentiate the following functions from first principles:

ecosx

Answer:

Let  fx=ecosx       fx+h=ecosx+h        ddxfx=limh0fx+h-fxh                             =limh0ecosx+h-ecosxh                             =limh0 ecosxecosx+h-cosx-1h                             =limh0 ecosxecosx+h-cosx-1cosx+h cosx×cosx+h-cosxh                             =ecosxlim             h0  cosx+h-cosxh ×limh0ecosx+h-cosx-1cosx+h-cosx                             =ecosxlimh0  cosx+h-cosxh                            limh0ex-1x=1                             =ecosxlimh0 -2sinx+h+x2sinx+h-x2h          cosA-cosB=-2sin A+B2sinA-B2                            =ecosxlimh0-sin2x+h21×sinh2h2                            =ecosxlimh0-sin2x+h21× lim       h0sinh2h2                           =ecosxlimh0-sin2x+h2                                         sinxx=1                          =ecosx-sinx                           =-sinxecosxHence, ddxecosx=-sinxecosx

Page No 10.17:

Question 5:

Differentiate the following functions from first principles:

e2x

Answer:

Let  fx=e2x  fx+h=e2x+h        ddxfx=limh0fx+h-fxh                             =limh0e2x+h-e2xh                             =limh0 e2xe2x+h-2x-1h                             =e2xlimh0 e2x+h-2x-12x+h-2x×limh02x+h-2xh                             =e2xlimh02x+h-2xh                                 limh0eh-1h=1                             =e2xlimh0 2x+h-2xh × 2x+h+2x2x+h+2x       Rationalising the numerator                            =e2xlimh02x+h-2xh2x+h+2x                           =e2xlimh02x+2h-2xh2x+h+2x                                                                =e2xlimh02hh2x+h+2x                           =e2xlimh022x+h+2x                          =e2x2xHence, ddxe2x=e2x2x

Page No 10.17:

Question 6:

Differentiate the following functions from first principles:

log cos x

Answer:

Let fx=log cosxfx+h=log cosx+hddxfx=limh0fx+h=fxh                     =limh0log cosx+h-log cosxh                     =limh0logcosx+hcosxh                        logA-logB=logAB                     =limh0log1+cosx+hcosx-1h                     =limh0log1+cosx+h-cosxcosxcosx+h-cosxcosx×limh0cosx+h-cosxcosx                     =1×limh0cosx+h-cosxcosx × h                    limx0log1+xx=1                    =limh0-2sinx+h+x2sinx+h-x2cosx × h                     =-2limh0sin2x+h2×sinh22cosx ×h2                     =-2sinx2cosx                            limx0sinxx=1                      =-tanxSo, ddxlog cosx=-tanx

Page No 10.17:

Question 7:

​Differentiate the following function from first principles:

ecot x
 

Answer:

  Let  fx=ecotxfx+h=ecotx+h  ddxfx=limh0fx+h-fxh                      =limh0ecotx+h-ecotxh                      =limh0ecotxecotx+h-cotx-1h                      =ecotxlimh0ecotx+h-cotx-1cotx+h-cotx×cotx+h-cotxh                      =ecotxlimh0cotx+h-cotxh×cotx+h+cotxcotx+h+cotx                            limx0ex-1x=1 and rationalizing the numerator                      =ecotxlimh0cotx+h-cotxhcotx+h+cotx                      =ecotxlimh0cotx+hcotx+1cotx-x-hhcotx+h+cotx                                                                     cotA-B=cotAcotB+1cotB-cotA                      =ecotxlimh0cotx+hcotx+1cot-h×hcotx+h+cotx                      =-ecotxlimh0cotx+hcotx+1htanhcotx+h+cotx                      =ecotx×cot2x+12cotx                                       limx0tanxx=1                      =-ecotx×cosec2x2cotx                                          1+cot2x=cosec2xddxecotx=-ecotx×cosec2x2cotx  

Page No 10.17:

Question 8:

Differentiate the following functions from first principles:

x
2ex
 

Answer:

 Let  fx=x2exfx+h=x+h2ex+h                 =limh0fx+h-fxh                 =limh0x+h2ex+h-x2exh                 =limh0x2ex+h-x2exh+2xhex+hh+h2ex+hh                 =limh0x2exex+h-x-1h+2xex+h+hex+h                 =limh0x2exeh-1h+2xex+h+hex+h                 =x2ex+2xex+0xex                        limx0ex-1x=1ddxx2ex=exx2+2x

Page No 10.17:

Question 9:

Differentiate the following functions from first principles:

 log cosec x

Answer:

 Let  fx=log cosecx         fx+h=log cosecx+hddxfx=limh0fx+h-fxh                     =limh0log cosecx+h-log cosecxh                     =limh0logcosecx+hcosecxh                     =limh0log1+sinxsinx+h-1h                     =limh0log1+sinx-sinx+hsinx+hsinx-sinx+hsinx+hsinx-sinx+hsinx+hh                     =limh02cosx+x+h2sinx-x-h2sinx+hh                             limx0log1+xx=1 and sinA-sinB=2cosA+B2sinA-B2                     =limh02cos2x+h2sinx+h -2sin-h2-h2                                 limx0sinxx=1                     =-cotxddxlog cosecx=-cotx

Page No 10.17:

Question 10:

Differentiate the following functions from first principles:

sin−1 (2x + 3)

Answer:

 Let fx=sin-12x+3        fx+h=sin-12x+h+3        fx+h=sin-12x+2h+3 ddxfx=limh0fx+h-fxh                      =limh0sin-12x+2h+3-sin-12x+3h                      =limh0sin-12x+2h+31-2x+32-2x+31-2x+2h+32h             sin-1x-sin-1y=sin-1x1-y2-y1-x2                      =limh0sin-1zz×zhwhere, z=2x+2h+31-2x+32-2x+31-2x+2h+32 and limh0sin-1hh=1                      =limh0zh                      =limh0 2x+2h+31-2x+32-2x+31-2x+2h+32 h                      =limh02x+2h+321-2x+32-2x+321-2x+2h+32h2x+2h+31-2x+32+2x+31-2x+2h+32           Rationalizing numerator                      =limh02x+32+4h2+4h2x+31-2x+32-2x+321-2x+32-4h2-4h2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh02x+32+4h2+4h2x+3-2x+34-4h22x+32-4h2x+33-2x+32+2x+34+4h22x+32+4h2x+33h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh04hh+2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =42x+32x+31-2x+32+2x+31-2x+32                      =42x+322x+31-2x+32                      =21-2x+32 ddxsin-12x+3=21-2x+32             



Page No 10.37:

Question 1:

Differentiate

sin (3x + 5)

Answer:

Let y=sin3x+5Differentiating y with respect to x we get,dydx=ddxsin3x+5        =cos3x+5ddx3x+5           using chain rule        =cos3x+5 ×3        =3cos3x+5So,ddxsin3x+5=3cos3x+5

Page No 10.37:

Question 2:

Differentiate

tan2x

Answer:

Let  y=tan2xDifferentiating with respect to x we get,dydx=2 tanxddxtanx             using chain rule        =2 tanx × sec2x           So, ddxtan2x=2 tanxsec2x 

Page No 10.37:

Question 3:

Differentiate

tan (x° + 45°)

Answer:

Let, y=tanx°+45°y=tanx+45π180Differentiating it with respect to x we get,dydx=ddxtanx+45π180        =sec2x+45π180×ddxx+45π180         Using chain rule        =π180sec2x°+45°So,      ddxtanx°+45°=π180sec2x°+45°

Page No 10.37:

Question 4:

Differentiate

sin (log x)

Answer:

Let  y=sinlogxDifferentiate it with respect to x we get,dydx=ddxsinlogx         =coslogxddxlogx        using chain rule         =1xcoslogxSo, ddxsinlogx=1xcoslogx

Page No 10.37:

Question 5:

Differentiate

esin x

Answer:

Let  y=esinxDifferentiate it with respect to x we get,dydx=ddxesinx        =esinxddxsinx              using chain rule        =esinx×cosxddxx       using chain rule        =esinx×cosx×12x        =cosx esinx2xSo, ddxesinx=cosx esinx2x

Page No 10.37:

Question 6:

Differentiate

etanx

Answer:

Let y=etanxDifferentiate it with respect to x we get,dydx=ddxetanx        =etanxddxtanx            using chain rule        =etanx × sec2xSo, ddxetanx= sec2xetanx

Page No 10.37:

Question 7:

Differentiate

sin2 (2x + 1)

Answer:

Let  y=sin22x+1Differentiate it with respect to x we get,dydx=ddxsin22x+1        =2sin2x+1ddxsin2x+1                        using chain rule        =2sin2x+1 cos2x+1 ddx2x+1        using chain rule         =4sin2x+1 cos2x+1        =2sin22x+1                                  sin2A=2sinAcosA        =2 sin4x+2So, ddxsin22x+1=2 sin4x+2

Page No 10.37:

Question 8:

Differentiate

log7 (2x − 3)

Answer:

Let y=log72x-3    y=log2x-3log7                      logab=logblogaDifferentiate it with respect to x we get,dydx=1log7ddxlog2x-3        =1log7×12x-3ddx2x-3    using chain rule        =22x-3log7Hence, ddxlog72x-3=22x-3log7

Page No 10.37:

Question 9:

Differentiate

tan 5x°

Answer:

Let y=tan5x°   y=tan5x×π180Differentiate it with respect to x we get,dydx=ddxtan5x×π180         =sec25x×π180ddx5x×π180        using chain rule         =5π180sec25x×π180         =5π180sec25x°Hence,  ddxtan5x°=5π180sec25x°

Page No 10.37:

Question 10:

Differentiate

2x3

Answer:

Let  y=2x3Differentiate it with respect to x we get,dydx=ddx2x3        =2x3×loge2ddxx3            using chain rule        =3x2 × 2x3×loge2Hence, ddx2x3=3x2 × 2x3loge2

Page No 10.37:

Question 11:

Differentiate

3ex

Answer:

Let y=3exDifferentiate it with respect to x we get,dydx=ddx3ex        =3exlog3ddxex           using chain rule        =ex×3exlog3So, ddx3ex=ex×3exlog3

Page No 10.37:

Question 12:

Differentiate

logx 3

Answer:

Let y=logx3   y=log3logx                      logab=logblogaDifferentiate it with respect to x we get,dydx=ddxlog3logx        =log3ddxlogx-1        =log3 × -1logx-2ddxlogx       using chain rule        =-log3logx2×1x        =-log3logx2×1x×1log3        =-1xlog3log3x2                             logbloga=logabSo, ddxlogx3=-1xlog3log3x2

Page No 10.37:

Question 13:

Differentiate

3x2+2x

Answer:

Let y=3x2+2xDifferentiate it with respect to x we get,dydx=ddx3x2+2x        =3x2+2x × loge3ddxx2+2x           using chain rule        =2x+23x2+2x loge3So, ddx3x2+2x=2x+23x2+2x loge3

Page No 10.37:

Question 14:

Differentiate

a2-x2a2+x2

Answer:

Let y=a2-x2a2+x2   y=a2-x2a2+x212Differentiate it with respect to x we get,dydx=ddxa2-x2a2+x212        =12a2-x2a2+x212-1×ddxa2-x2a2+x2                                                       Using chain rule        =12a2-x2a2+x2-12×a2+x2ddxa2-x2-a2-x2ddxa2+x2a2+x22         =12a2+x2a2-x212-2xa2+x2-2xa2-x2a2+x22        =12a2+x2a2-x212-2xa2-2x3-2xa2+2x3a2+x22        =12a2+x2a2-x212-4xa2a2+x22        =-2xa2a2-x2a2+x232So,  ddxa2-x2a2+x2=-2a2xa2-x2a2+x232

Page No 10.37:

Question 15:

Differentiate

3x log x

Answer:

Let  y=3x logxDifferentiate it with respect to x we get,dydx=ddx3x logx        =3x logx×loge3ddxx logx                             Using chain rule        =3x logx×loge3xddxlogx+logxddxx          =3x logx×loge3xx+logx         =3x logx1+logx×loge3So, ddx3x logx=3x logx1+logxloge3

Page No 10.37:

Question 16:

Differentiate

1+sin x1-sin x

Answer:

Let  y=1+sinx1-sinxDifferentiate it with respect to x we get,dydx=ddx1+sinx1-sinx12       =121+sinx1-sinx12-1ddx1+sinx1-sinx       =121-sinx1+sinx121-sinxcosx-1+sinx-cosx1-sinx2       =121-sinx121+sinx12cosx-cosx sinx+cosx+sinx cosx1-sinx2       =12×2cosx1+sinx1-sinx32       =cosx1+sinx1-sinx32       =cosx1+sinx1-sinx1-sinx       =cosx1-sin2x×1-sinx        =cosxcosx1-sinx                        Using 1-sin2x=cos2x        =11-sinx×1+sinx1+sinx        =1+sinx1-sin2x         =1+sinxcos2x         =1cosx1cosx+sinxcosx         =secxsecx+tanxHence, dydx=secxsecx+tanx

Page No 10.37:

Question 17:

Differentiate

1-x21+x2

Answer:

Let  y=1-x21+x2     y=1-x21+x212Differentiate it with respect to x we get,dydx=ddx1-x21+x212        =121-x21+x212-1×ddx1-x21+x2           Using chain rule        =121-x21+x2-12×1+x2ddx1-x2-1-x2ddx1+x21+x22          =121+x21-x212-2x1+x2-2x1-x21+x22         =121+x21-x212-2x-2x3-2x+2x31+x22          =121+x21-x212-4x1+x22           =-2x1-x21+x232

Page No 10.37:

Question 18:

Differentiate

(log sin x)2

Answer:

Let y=log sinx2Differentiate with respect to x we get,dydx=ddxlog sinx2        =2log sinxddxlog sinx        =2log sinx×1sinxddxsinx        =2log sinx×1sinx×cosx        =2log sinxcotxSo, ddxlog sinx2=2log sinxcotx

Page No 10.37:

Question 19:

Differentiate

1+x1-x

Answer:

Let  y=1+x1-x     y=1+x1-x12Differentiate it with respect to x we get,dydx=ddx1+x1-x12        =121+x1-x12-1×ddx1+x1-x           Using chain rule        =121+x1-x-12×1-xddx1+x-1+xddx1-x1-x2   Using quotient rule        =121-x1+x121-x1-1+x-11-x2         =121-x1+x121-x+1+x1-x2          =121-x121+x12×21-x2           =11+x1-x32So, ddx1+x1-x=11+x1-x32

Page No 10.37:

Question 20:

Differentiate

sin 1+x21-x2

Answer:

Let y=sin1+x21-x2Differentiate it with respect to x we get,dydx=ddxsin1+x21-x2         =cosx1+x21-x2ddx1+x21-x2                   Using chain rule         =cosx1+x21-x21-x2ddx1+x2-1+x2ddx1-x21-x22       Using quotient rule         =cosx1+x21-x21-x22x-1+x2-2x1-x22         =cosx1+x21-x22x-2x3+2x+2x31-x22        =4x1-x22cosx1+x21-x2So,ddxsin1+x21-x2=4x1-x22cosx1+x21-x2     

Page No 10.37:

Question 21:

Differentiate

e3 x cos 2x

Answer:

Let  y=e3x cos2xDifferentiate it with respect to x we get,dydx=ddxe3x cos2x        =e3x×ddxcos2x+cos2xddxe3x               Using product rule        =e3x×-sin2xddx2x+cos2xe3xddx3x   Using chain rule        =-2e3x sin2x+3e3x cos2x        =e3x3 cos2x-2 sin2xSo,ddxe3x cos2x=e3x3 cos2x-2 sin2x

Page No 10.37:

Question 22:

Differentiate

sin (log sin x)

Answer:

Let y=sinlog sinxDifferentiate it with respect to x we get,dydx=ddxsinlog sinx        =coslog sinxddxlog sinx                Using chain rule        =coslog sinx×1sinxddx sinx          Using chain rule        =coslog sinxcosxsinx        =coslog sinx cotxHence, ddxsinlog sinx=coslog sinx cotx

Page No 10.37:

Question 23:

Differentiate

etan 3 x

Answer:

Let y=etan3xDifferentiate it with respect to x we get,dydx=ddxetan3x       =etan3xddxtan3x                =etan3xsec23x ×ddx3x       =etan3xsec23x ×3So, ddxetan3x=3etan3xsec23x

Page No 10.37:

Question 24:

Differentiate

ecot x

Answer:

Let  y=ecotx   y=ecotx12Differentiate it with respect to x we get,dydx=ddxecotx12        =ecotx12×ddxcotx12             Using chain rule        =ecotx×12cotx12-1ddxcotx        =-ecotx×cosec2x2cotxSo, ddxecotx=-ecotx×cosec2x2cotx

Page No 10.37:

Question 25:

Differentiate

log sin x1+cos x

Answer:

Let y=logsinx1+cosxDifferentiate it with respect to x, we getdydx=ddxlogsinx1+cosx        =1sinx1+cosx×ddxsinx1+cosx                   Using chain rule        =1+cosxsinx1+cosxddxsinx-sinxddx1+cosx1+cosx2         Using quotient rule        =1+cosxsinx1+cosxcosx-sinx-sinx1+cosx2        =1+cosxsinxcosx+cos2x+sin2x1+cosx2        =1+cosxsinx1+cosx1+cosx2        =1sinx        =cosecxSo, ddxlogsinx1+cosx=cosecx

Page No 10.37:

Question 26:

Differentiate

log 1-cos x1+cos x

Answer:

Let  y=log1-cosx1+cosx   y=log1-cosx1+cosx12   y=12log1-cosx1+cosx                        using logab=blogaDifferentiate it with respect to x we get,dydx=ddx12log1-cosx1+cosx        =12×11-cosx1+cosx×ddx1-cosx1+cosx          Using chain rule         =121+cosx1-cosx1+cosxddx1-cosx-1-cosxddx1+cosx1+cosx2              Using quotient rule         =121+cosx1-cosx1+cosxsinx-1-cosx-sinx1+cosx2         =121+cosx1-cosxsinx+sinx cosx+sinx-sinx cosx1+cosx2         =121+cosx1-cosx2sinx1+cosx2          =sinx1-cosx1+cosx          =sinx1-cos2x          =sinxsin2x                                     =1sinx         =cosecxSo, ddxlog1-cosx1+cosx=cosecx

Page No 10.37:

Question 27:

Differentiate

tan esin x

Answer:

Let y=tanesinxDifferentiate it with respect to x we get, dydx=ddxtanesinx         =sec2esinxddxesinx                Using chain rule         =sec2esinx×esinx×ddxsinx        =cosxsec2esinx×esinxSo, ddxtanesinx=cosxsec2esinxesinx  

Page No 10.37:

Question 28:

Differentiate

log x+x2+1

Answer:

Let  y=logx+x2+1Differentiate with respect to x we get,dydx=ddxlogx+x2+1        =1x+x2+1ddxx+x2+112                  Using chain rule        =1x+x2+11+12x2+112-1ddxx2+1        =1x+x2+11+12x2+1×2x        =1x+x2+1x2+1+xx2+1        =1x2+1So,  ddxlogx+x2+1=1x2+1

Page No 10.37:

Question 29:

Differentiate

ex log xx2

Answer:

Let   y=exlogxx2Differentiate with respect to x we get,dydx=x2ddxexlogx-exlogxddxx2x22                Using quotient rule        =x2exddxlogx+logxddxex-ex logx×2xx4            Using product rule        =x2exx+ex logx-2xex logxx4        =x2ex1+xlogxx-2xex logxx4        =xex1+xlogx-2logxx4        =xexx31x+x logxx-2logxx        =exx-21x+logx-2xlogxSo, ddxexlogxx2  = exx-21x+logx-2xlogx     

Page No 10.37:

Question 30:

Differentiate

log cosec x-cot x

Answer:

Let y=log cosecx-cotxDifferentiate it with respect to x we get,dydx=ddxlog cosecx-cotx        =1cosecx-cotxddxcosecx-cotx        =1cosecx-cotx×-cosecx cotx+cosec2x        =cosecxcosecx-cotxcosecx-cotx        =cosecxSo,ddxlog cosecx-cotx=cosecx

Page No 10.37:

Question 31:

Differentiate

e2x+e-2xe2x-e-2x

Answer:

Let  y=e2x+e-2xe2x-e-2x
Differentiate with respect to x we get,
dydx=ddxe2x+e-2xe2x-e-2x         =e2x-e-2xddxe2x+e-2x-e2x+e-2xddxe2x-e-2xe2x-e-2x2              Using quotient rule and chain rule         =e2x-e-2xe2xddx2x+e-2xddx-2x-e2x+e-2xe2xddx2x-e-2xddx-2xe2x-e-2x2         =e2x-e-2x2e2x-2e-2x-e2x+e-2x2e2x+2e-2xe2x-e-2x2         =2e2x-e-2x2-2e2x+e-2x2e2x-e-2x2         =2e4x+e-4x-2e2xe-2x-e4x-e-4x-2e2xe-2xe2x-e-2x2         =-8e2x-e-2x2So, ddxe2x+e-2xe2x-e-2x=-8e2x-e-2x2

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Question 32:

Differentiate

log x2+x+1x2-x+1

Answer:

Let  y=logx2+x+1x2-x+1

Differentiate with respect of x we get,

dydx=ddxlogx2+x+1x2-x+1        =1x2+x+1x2-x+1ddxx2+x+1x2-x+1             Using chain rule and quotient rule        =x2-x+1x2+x+1x2-x+1ddxx2+x+1-x2+x+1ddxx2-x+1x2-x+12        =x2-x+1x2+x+1x2-x+12x+1-x2+x+12x-1x2-x+12        =x2-x+1x2+x+12x3-2x2+2x+x2-x+1-2x3-2x2-2x+x2+x+1x2-x+12        =-4x2+2x2+2x2+x+1x2-x+1        =-4x2+2x2+2x2+12-x2        =-2x2-1x4+1+2x2-x2        =-2x2-1x4+x2+1So, ddxlogx2+x+1x2-x+1=-2x2-1x4+x2+1

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Question 33:

Differentiate

tan-1 ex

Answer:

Let y=tan-1ex
Differentiate it with respect to x we get,
dydx=ddxtan-1ex       =11+ex2ddxex                 Using chain rule       =11+e2x×ex       =ex1+e2xSo, ddxtan-1ex=ex1+e2x

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Question 34:

Differentiate

esin-1 2x

Answer:

Let  y=esin-12x
Differentiate it with respect to x we get,
dydx=ddxesin-12x       =esin-12x×ddxsin-12x             Using chain rule       =esin-12x×11-2x2ddx2x       =2esin-12x1-4x2So, ddxesin-12x=2esin-12x1-4x2

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Question 35:

Differentiate

sin 2 sin-1 x

Answer:

Let y=sin2 sin-1x
Differentiate it with respect to x we get,
dydx=ddxsin2sin-1x       =cos2 sin-1xddx2 sin-1x            Using chain rule       =cos2 sin-1x×211-x2       =2cos2 sin-1x1-x2So,ddxsin2sin-1x=2cos2 sin-1x1-x2

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Question 36:

Differentiate

etan-1 x

Answer:

Let y=etan-1x
Differentiate it with respect to x we get,
dydx=ddxetan-1x       =etan-1xddxtan-1x             Using chain rule       =etan-1x×11+x2ddxx       =etan-1x1+x×12x       =etan-1x2x1+xSo,  ddxetan-1x=etan-1x2x1+x

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Question 37:

Differentiate

tan-1x2

Answer:

Let  y=tan-1x2y=tan-1x212
Differentiate it with respect to x we get,
dydx=ddxtan-1x212       =12tan-1x212-1ddxtan-1x2                 Using chain rule       =12tan-1x2-12×11+x22×ddxx2       =444+x2tan-1x2       =14+x2tan-1x2So, ddxtan-1x2=14+x2tan-1x2

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Question 38:

Differentiate

log tan-1 x

Answer:

Let y=logtan-1x
Differentiate it with respect to x we get,
dydx=ddxlogtan-1x       =1tan-1x×ddxtan-1x         Using chain rule       =11+x2tan-1xSo, ddxlogtan-1x=11+x2tan-1x

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Question 39:

Differentiate

2x cos xx2+32

Answer:

Let  y=2x cosxx2+32
Differentiate it with respect to x we get,
dydx=ddx2x cosxx2+32       =x2+32ddx2x cosx-2x cosxddxx2+32x2+322          Using quotient rule       =x2+322xddxcosx+cosxddx2x-2x cosx2x2+3ddxx2+3x2+34       Using Product rule and chain rule       =x2+32-2x sinx+cosx2x loge2-22x cosxx2+32xx2+34       =2xx2+3x2+3cosx  loge2-sinx-4x cosxx2+34       =2xx2+32cosx loge2-sinx-4x cosxx2+3So, ddx2x cosxx2+32=2xx2+32cosx loge2-sinx-4x cosxx2+3

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Question 40:

Differentiate

x sin 2x+5x+kk+ tan2 x3

Answer:

Let y=x sin2x+5x+kk+tan2x3

Differentiate it with respect to x we get,

dydx=ddxx sin2x+5x+kk+tan6x       =ddxx sin2x+ddx5x+ddxkk+ddxtan6x       =xddxsin2x+sin2xddxx+5x loge5+0+6tan5x×ddxtanx         Using product rule and chain rule       =x cos2xddx2x+sin2x+5xloge5+6tan5x sec2x       =2x cos2x+sin2x+5x loge5+6tan5x sec2xSo, ddxx sin2x+5x+kk+tan2x3=2xcos2x+sin2x+5x loge5+6tan5x sec2x

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Question 41:

Differentiate

log 3x+2-x2 log 2x-1

Answer:

Let  y=log3x+2-x2log2x-1
Differentiate it with respect to x we get,

dydx=ddxlog3x+2-x2log2x-1        =ddxlog3x+2-ddxx2log2x-1        =13x+2ddx3x+2-x2ddxlog2x-1+log2x-1ddxx2            Using product rule and chain rule        =33x+2-2x22x-1-2x log2x-1So, ddxlog3x+2-x2log2x-1=33x+2-2x22x-1-2x log2x-1     

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Question 42:

Differentiate

3x2 sin x7-x2

Answer:

Let  y=3x2 sinx7-x2
Differentiate it with respect to x we get,
dydx=ddx3x2 sinx7-x212        =7-x212×ddx3x2 sinx-3x2 sinxddx7-x2127-x2122            Using quotient rule, chain rule and product rule        =7-x212× 3x2ddxsinx+sinxddxx2-3x2sinx × 127-x2×ddx7-x27-x2        =7-x2123x2cosx+2x sinx-3x2 sinx × 127-x2-12-2x7-x2        =7-x212× 3x2 cosx+2x sinx7-x2+3x3sinx7-x2-127-x2        =6x sinx+3x2cosx7-x2+3x3 sinx7-x232So, ddx3x2 sinx7-x2=6x sinx+3x2cosx7-x2+3x3 sinx7-x232

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Question 43:

Differentiate

sin2 log 2x+3

Answer:

Let y=sin2log2x+3
dydx=ddxsin2log2x+3             =2 sinlog2x+3ddxsinlog2x+3          Using chain rule             =2sinlog2x+3 coslog2x+3ddxlog2x+3             =sin2log2x+3×12x+3ddx2x+3      2sinA cosA=sin2A             =sin2log2x+322x+3So, ddxsin2log2x+3=sin2 log2x+322x+3

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Question 44:

Differentiate

ex log sin 2x

Answer:

Let y=exlog sin2x
Differentiate it with respect to x we get,

dydx=ddxex log sin2x       =exddxlog sin2x+log sin2xddxex          Using product rule and chain rule       =ex1sin2xddxsin2x+log sin2xex       =exsin2xcos2x ddx2x+ex log sin2x       =2cos2xexsin2x+ex log sin2x       =2excot2x+exlog sin2xSo, ddxex log sin2x=2excot2x+exlog sin2x

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Question 45:

Differentiate

x2+1+x2-1x2+1-x2-1

Answer:

We have, x2+1+x2-1x2+1-x2-1By rationalising we get,x2+1+x2-1x2+1-x2-1×x2+1+x2-1x2+1+x2-1=x2+1+x2-12x2+12-x2-12=x2+12+x2-12+2x2+1x2-1x2+1-x2+1=x2+1+x2-1+2x4-12=2x2+2x4-12=x2+x4-1Now, Let y=x2+x4-1

Differentiate it with respect to x we get,

dydx=ddxx2+x4-1       =2x+12x4-1×ddxx4-1       =2x+12x4-1×4x3       =2x+2x3x4-1

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Question 46:

Differentiate

log x+2+x2+4x+1

Answer:

Let y=logx+2+x2+4x+1
Differentiate it with respect to x we get,
dydx=ddxlogx+2+x2+4x+1       =1x+2+x4+4x+1ddxx+2+x2+4x+112        Using chain rule       =1x+2+x4+4x+1×1+0+12x2+4x+1-12ddxx2+4x+1       =1+2x+42x2+4x+1x+2+x4+4x+1       =x2+4x+1+x+2x+2+x2+4x+1×x2+4x+1       =1x2+4x+1So, ddxlogx+2+x2+4x+1=1x2+4x+1

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Question 47:

Differentiate

sin-1 x44

Answer:

Let y=sin-1x44
Differentiate it with respect to x we get,
dydx=ddxsin-1x44       =4sin-1x43ddxsin-1x4         Using chain rule       =4sin-1x4311-x42ddxx4            Using chain rule       =4sin-1x434x31-x8       =16x3sin-1x431-x8So,  ddxsin-1x44=16x3sin-1x431-x8

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Question 48:

Differentiate

sin-1 xx2+a2

Answer:

Let y=sin-1xx2+a2
Differentiate it with respect to x we get,
dydx=ddxsin-1xx2+a2         =11-xx2+a22×ddxxx2+a2             Using chain rule and quotient rule         =11-xx2+a22×x2+a212ddxx-xddxx2+a212x2+a2122         =x2+a2x2+a2-x2x2+a2- x2x2+a2ddxx2+a2x2+a2         =x2+a2ax2+a2x2+a2-x2x2+a2×2x         =x2+a2ax2+a2x2+a2-x2x2+a2         =a2ax2+a2         =ax2+a2So,ddxsin-1xx2+a2=ax2+a2

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Question 49:

Differentiate

ex sin xx2+23

Answer:

Let  y=exsinxx2+23
Differentiate it with respect to we get,
dydx=x2+23ddxexsinx-exsinxddxx2+23x2+232          Using quotient rule         =x2+23excosx+sinxex-ex sinx 3x2+222xx2+26    Using product rule         =x2+23excosx+exsinx-6xex sinxx2+22x2+26         =x2+22x2+2excosx+exsinx-6xex sinxx2+26          =x2+2excosx+exsinx-6xex sinxx2+24         =ex sinx+ex cosxx2+23-6xex sinxx2+24So, dydx=ex sinx+ex cosxx2+23-6xex sinxx2+24

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Question 50:

Differentiate

3 e-3x log 1+x

Answer:

Let y=3e-3x log1+x
Differentiate it with respect to x we get,

dydx=3ddxe-3x log1+x                  =3e-3x11+x+log1+x-3e-3x   Using product rule and chain rule        =3e-3x1+x-3e-3xlog1+x        =3e-3x11+x-3 log1+xSo, ddx3e-3x log1+x=3e-3x11+x-3 log1+x

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Question 51:

Differentiate

x2+2cos x

Answer:

Let y=x2+2cosx
Differentiate it with respect to x we get,
dydx=cosxddxx2+2-x2+2ddxcosxcosx2       Using quotient rule and chain rule         =2xcosx-x2+2-12sinxcosxcosx         =2xcosx+x2+2sinx2cosxcosx         =4x cosx+x2+2sinx2cosx32         =2xcosx+12x2+2sinxcosx32        =1cosx2x+12x2+2sinxcosx        =1cosx2x+x2+2tanx2So, ddxx2+2cosx=1cosx2x+x2+2tanx2



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Question 52:

Differentiate

x2 1-x23cos 2x

Answer:

Let y=x21-x23cos 2x
dydx=cos2xddxx21-x23-x21-x23ddxcos2xcos22x             =cos2xx2ddx1-x23+1-x23ddxx2-x21-x23-2sin2xcos22x         =cos2x-6x31-x22+1-x232x+2x21-x23sin2xcos22x         =2x1-x22cos2x-6x31-x22cos2x+2x21-x23sin2xcos22x         =2x1-x2sec2x1-4x2+x1-x2tan2xSo, ddxx21-x23cos2x=2x1-x2sec2x1-4x2+x1-x2tan2x

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Question 53:

logcotπ4+x2

Answer:

logcotπ4+x2=-cosec2π4+x22cotπ4+x2=-12cosπ4+x2sinπ4+x2=-1sinπ2+x=-1cosx=-secx

Disclaimer: The answer given at the back of the exercise in RD Sharma is incorrect.

 

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Question 54:

Differentiate

eax sec x tan 2x

Answer:

Let y=eax secx tan2x
Differentiate it with respect to x,
dydx=ddxeax secx tan2x        =eaxddxsecx tan2x+secx tan2xddxeax                 =eaxsecx tanx tan2x+2sec22xsecx+aeax secx tan2x           =eaxsecx tanx tan2x+2sec22xsecx+aeax secx tan2x        =aeax secx tan2x+eaxsecx tanx tan2x+2eax secxsec22x        =eax secxa tan2x+tanx tan2x+2sec22xSo,ddxeax secx tan2x=eax secxa tan2x+tanx tan2x+2sec22x

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Question 55:

Differentiate

log cos x2

Answer:

Let y=log cosx2
Differentiating with respect to x,
dydx=ddxlogcosx2        = -2x sinx2cosx2                           =-2x tanx2So, ddxlogcosx2=-2x tanx2

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Question 56:

Differentiate

cos log x2

Answer:

Let y=cos log x2
Differentiating with respect to x,
dydx=ddxcoslog x2        =-sinlog x2ddxlog x2              =-sinlog x22logxx                         =-2logx sinlog x2 xSo, ddxcoslog x2=-2logx sinlog x2 x

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Question 57:

Differentiate

log x-1x+1

Answer:

Let y=log x-1x+1  y=log x-1x+112  y=12log  x-1x+1  y=12logx-1-logx+1
Differentiate it with respect to x
dydx=12ddxlogx-1-ddxlogx+1        =121x-1-1x+1        =122x2-1        =1x2-1So,dydx=1x2-1

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Question 58:

If y=log x-1-x+1, show that dydx=-12x2-1

Answer:

Here,  y=logx-1-x+1
Differentiate it with respect to x we get,
dydx=ddxlogx-1-x+1        =1x-1-x+1ddxx-1-x+1           Using chain rule        =1x-1-x+1ddxx-1-ddxx+1        =1x-1-x+112x-1-12-12x+1-12        =121x-1-x+11x-1-1x+1        =121x-1-x+1-x-1-x+1x-1x+1        =-121x-1x+1        =-12x2-1So,dydx=-12x2-1

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Question 59:

If y=x+1+x-1, prove that x2-1dydx=12y

Answer:

We have, y=x+1+x-1
Differentiating with respect to x,
dydx=ddxx+1+ddxx-1dydx=12x+1-12+12x-1-12dydx=121x+1+1x-1dydx=12x-1+x+1x+1x-1dydx=12yx2-1x2-1dydx=12yHence proved

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Question 60:

If y=xx+2, prove that xdydx=1-y y

Answer:

We have, y=xx+2
Differentiating with respect to x,

dydx=ddxxx+2dydx=x+2ddxx-xddxx+2x+22       dydx=x+2-xx+22dydx=x+2x+22-xx+22dydx=1x+2-xy2x2                    x+2=xydydx=yx-y2xdydx=1xy1-yxdydx=1-yy Hence proved

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Question 61:

If y=log x+1x, prove that dydx=x-12x x+1

Answer:

We have, y=logx+1x
Differentiate it with respect to x,
dydx=ddxlogx+1x        =1x+1xddxx+1x                 =xx+112x-12xx        =12xx+1x-1xx        =x-12xx+1So, dydx=x-12xx+1

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Question 62:

If y=log 1+tan x1-tan x, prove that dydx=sec 2x.

Answer:

​Let y=log1+tanx1-tanxy=log1+tanx1-tanx12y=12log1+tanx1-tanxy=12log1+tanx-log1-tanxdydx=12ddxlog1+tanx-ddxlog1-tanx            =1211+tanx×ddx1+tanx-11-tanx×ddx1-tanx            =1211+tanx0+sec2x-11-tanx0-sec2x            =12sec2x1+tanx+sec2x1-tanx            =12sec2x1-tanx+1+tanx1-tan2x            =12sec2x21-tan2x            =sec2x1-tan2x            =1+tan2x1-tan2x            =11-tan2x1+tan2x            =1cos2x            =sec2x

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Question 63:

If y=x+1x, prove that 2 xdydx=x-1x

Answer:

We have,  y=x+1x
Differentiate with respect to x,
dydx=ddxx+1xdydx=ddxx+ddx1xdydx=12x+-12xxdydx=12x-12xxdydx=x-12xx2xdydx=x-1x2xdydx=xx-1x2xdydx=x-1x

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Question 64:

If y=x sin-1 x1-x2, prove that 1-x2 dydx=x+yx.

Answer:

We have, y=x sin-1x1-x2
Differentiating with respect to x,
dydx=ddxx sin-1x1-x2dydx=1-x2ddxx sin-1x-x sin-1xddx1-x21-x22            dydx=1-x2xddxsin-1x+sin-1xddxx-x sin-1x121-x2ddx1-x21-x2     dydx=1-x2x1-x2+sin-1x-x sin-1x-2x21-x21-x2dydx=x+1-x2sin-1x+x2sin-1x1-x21-x21-x2dydx=x+1-x2sin-1x1+x2sin-1x1-x21-x2dydx=x+1-x2sin-1x+x2sin-1x1-x21-x2dydx=x+sin-1x-x2 sin-1x+x2sin-1x1-x21-x2dydx=x+sin-1x1-x21-x2dydx=x+yx                                   y=x sin-1x1-x2

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Question 65:

If y=ex-e-xex+e-x, prove that dydx=1-y2

Answer:

We have,  y=ex-e-xex+e-x
Differentiating with respect to x,
dydx=ddxex-e-xex+e-x        =ex+e-xddxex-e-x-ex-e-xddxex+e-xex+e-x2             =ex+e-xex-e-xddx-x-ex-e-xex+e-xddx-xex+e-x2        =ex+e-xex+e-x-ex-e-xex-e-xex+e-x2        =ex+e-x2-ex-e-x2ex+e-x2        =1-ex-e-x2ex+e-x2        =1-ex-e-xex+e-x2        =1-y2So, dydx=1-y2

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Question 66:

If y=x-1 log x-1-x+1 log x+1, prove that dydc=log x-11+x

Answer:

We have, y=x-1 logx-1-x+1 logx+1
Differentiating with respect to x,
dydx=ddxx-1 logx-1-x+1 logx+1        =x-1ddxlogx-1+logx-1ddxx-1-x+1ddxlogx+1+logx+1ddxx+1                     =x-1×1x-1ddxx-1+logx-1×1-x+1×1x+1×ddxx+1+logx+11        =1+logx-1-1+logx+1        =logx-1-logx+1        =logx-1x+1      So, dydx=logx-1x+1              

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Question 67:

If y=ex cos x, prove that dydx=2 ex·cos x+π4

Answer:

We have, y=ex cosx
Differentiating with respect to x,
dydx=ddxex cosx        =exddxcosx+cosxddxex              =ex-sinx+excosx        =excosx-sinx        =2ex cosx2-sinx2               Multiplying and dividing by 2        =2excosπ4cosx-sinπ4sinx        =2ex cosx+π4So, dydx=2ex cosx+π4

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Question 68:

If y=12 log 1-cos 2x1+cos 2x, prove that dydx=2 cosec 2x

Answer:

We have,  y=12log1-cos2x1+cos2xy=12log2sin2x2cos2x          y=12logtan2xy=22log tanx                     y=log tanx
Differentiate with respect to x,
dydx=log tanx        =1tanx×ddxtanx                   =sec2xtanx        =1cos2x × sinxcosx        =1sinx cosx        =22sinx cosx        =2sin2x                           So,dydx=2cosec 2x

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Question 69:

If y=x sin-1 x+1-x2, prove that dydx=sin-1 x

Answer:

We have, y=x sin-1x+1-x2
Differentiate it with respect to x,
dydx=ddxx sin-1x+1-x2        =ddxx sin-1x+ddx1-x2        =x ddxsin-1x+sin-1xddxx+121-x2ddx1-x2                   =x1-x2+sin-1x-2x21-x2        =x1-x2+sin-1x-x1-x2        =sin-1x

Page No 10.38:

Question 70:

If y=x2+a2, prove that ydydx-x=0

Answer:

We have, y=x2+a2
Squaring both sides we get,
y2=x2+a22ydydx=ddxx2+a22ydydx=2x2ydydx=2xydydx=xydydx-x=0Hence proved

Page No 10.38:

Question 71:

If y=ex+e-x, prove that dydx=y2-4

Answer:

We have, y=ex+e-x
Differentiate it with respect to x,
dydx=ddxex+e-x        =ddxex+ddxe-x        =ex+e-xddx-x               Using chain rule        =ex+e-x-1        =ex-e-x        =ex+e-x2-4ex×e-x            a-b=a+b2-4ab        =y2-4                           ex+e-x=y

Page No 10.38:

Question 72:

If y=a2-x2, prove that ydydx+x=0

Answer:

We have, y=a2-x2

Squaring both sides we get,y2=a2-x2Differentiating both sides w.r.t x, we get,2ydydx=ddxa2-x22ydydx=0-2x     ydydx=-xydydx+x=0  Hence proved

Page No 10.38:

Question 73:

If xy=4, prove that xdydx+y2=3 y

Answer:

We have, xy=4y=4x
Differentiate it with respect to x,
dydx=ddx4xdydx=4ddxx-1dydx=4-1× x-1-1dydx=4-1x2dydx=-4x2dydx=-44y2               x=4ydydx=-4y216dydx=-y244dydx=-y24dydx=3y2-4y24dydx+4y2=3y24dydx+y2=3y2
Dividing both side by x,
4xdydx+y2=3y2xydydx+y2=3y2x           xdydx+y2=3y2yxdydx+y2=3y

Page No 10.38:

Question 74:

Prove that ddx x2a2-x2+a22 sin-1xa=a2-x2

Answer:

ddxx2a2-x2+a22sin-1xa=a2-x2LHS=ddxx2a2-x2+a22sin-1xa        =ddxx2a2-x2+ddxa22sin-1xa        =12xddxa2-x2+a2-x2ddxx+a22×11-xa2×ddxxa               =12x × 12a2-x2ddxa2-x2+a2-x2+a22×1a2-x2a2×1a        =12x-2x2a2-x2+a2-x2+a22aa2-x2×1a        =12-2x2+2a2-x22a2-x2+a22a2-x2        =122a2-2x22a2-x2+a22a2-x2        =a2-2x22a2-x2+a22a2-x2        =a2-2x2+a22a2-x2        =2a2-2x22a2-x2        =2a2-x22a2-x2        =a2-x2a2-x2        =a2-x2=RHSHence proved



Page No 10.62:

Question 1:

Differentiate

cos-1 2x1-x2, 12<x<1

Answer:

Let,      y=cos-12x1-x2Put     x=cosθ           y=cos-12cosθ1-cos2θ          y=cos-12cosθ sinθ          y=cos-1sin2θ                       Since, sin2θ=2sinθcosθ          y=cos-1cosπ2-2θ        ...iNow,       12<x<112<cosθ<1  0<θ<π4  0<2θ<π2  0>-2θ>-π2π2>π2-2θ>0Hence, from equation iy=π2-2θ              Since, cos-1cosθ=θ, if θ0,πy=π2-2cos-1x    Since, x=cosθ differentiating it with respect to x,dydx=ddxπ2-2ddxcos-1xdydx=0-2-11-x2dydx=21-x2

Page No 10.62:

Question 2:

Differentiate

cos-1 1+x2, -1<x<1

Answer:

Let,       y=cos-11+x2Put       x=cos2θ             y=cos-11+cos 2θ2             y=cos-12cos2θ2             y=cos-1cosθ          ...iHere,  -1<x<1        -1<cos2θ<1          0<2θ<π        0<θ<π2So, from equation i      y=θ                            since ,cos-1cosθ=θ, ifθ0,π  y=12cos-1x             Since , x=cos2θDifferentiating it with respect to x,dydx=-121-x2



Page No 10.63:

Question 3:

Differentiate

sin-1 1-x2, 0<x<1

Answer:

Let,   y=sin-11-x2Put   x=cos 2θ         y=sin-11-cos2θ2         y=sin-12 sin2θ2         y=sin-1sinθ             ...iHere,  0<x<1       0<cos 2θ<1        0< 2θ<π2        0< θ<π4So, from equation i,       y=θ                        Since, sin-1sinθ=θ, if θ-π2,π2      y=12cos-1x            Since, x=cos 2θDifferentiating it with respect to x,dydx=-121-x2

Page No 10.63:

Question 4:

Differentiate

sin-1 1-x2, 0<x<1

Answer:

Let,   y=sin-11-x2Put,   x=cos θ         y=sin-11-cos2θ         y=sin-1sinθ           ...iHere,  0<x<1       0<cos θ<1        0< θ<π2So, from equation i,       y=θ                        Since, sin-1sinθ=θ, if θ-π2,π2      y=cos-1x            Since, x=cos θDifferentiating it with respect to x,dydx=-11-x2

Page No 10.63:

Question 5:

Differentiate

tan-1 xa2-x2, -a<x<a

Answer:

Let,   y=tan-1xa2-x2Put   x=a sinθ         y=tan-1a sinθa2-a2sin2θ         y=tan-1a sinθa21-sin2θ              y=tan-1a sinθa cosθ              y=tan-1tanθ      ...iHere,  -a<x<a       -1<xa<1    sin-π2<  sinθ<sinπ2      x=a sinθ        -π2< θ<π2So, from equation i,       y=θ                        Since, tan-1tanθ=θ, if θ-π2,π2      y=sin-1xa            Since, x=a sinθDifferentiating it with respect to x,Using chain rule,dydx=11-xa2ddxxadydx=aa2-x2×1adydx=1a2-x2

Page No 10.63:

Question 6:

Differentiate

sin-1 xx2+a2

Answer:

Let,   y=sin-1xx2+a2put   x=a tanθ     y=sin-1a tanθa2tan2θ+a2    y=sin-1a tanθa2tan2θ+1   y=sin-1a tanθa secθ      y=sin-1sinθ       y=θ     y=tan-1xa                  since, x=a tanθ   Differentiating it with respect to x using chain rule,  dydx=11+xa2ddxxadydx=a2a2+x2×1adydx=aa2+x2

Page No 10.63:

Question 7:

Differentiate

sin-1 2x2-1, 0<x<1

Answer:

Let,   y=sin-12x2-1Put   x=cos θ         y=sin-12cos2θ-1         y=sin-1cos2θ         y=sin-1sinπ2-2θ      ...i      Here,  0<x<1       0<cos θ<1        0< θ<π2        0< 2θ<π        0> -2θ>-π        π2>π2-2θ>-π2     -π2<π2-2θ<π2So, from equation i,       y=π2-2θ                        Since, sin-1sinθ=θ, if θ-π2,π2      y=π2-2cos-1x            Since, x=cos θDifferentiating it with respect to x,    dydx=0-2ddxcos-1xdydx=-2-11-x2dydx=21-x2

Page No 10.63:

Question 8:

Differentiate

sin-1 1-2x2, 0<x<1

Answer:

Let,   y=sin-11-2x2put   x=sin θ      y=sin-11-2sin2θ     y=sin-1cos2θ    y=sin-1sinπ2-2θ       ...i      Here,  0<x<1       0<sin θ<1        0< θ<π2        0< 2θ<π        0> -2θ>-π        π2>π2-2θ>π2-π        π2>π2-2θ>-π2     -π2<π2-2θ<π2So, from equation i,       y=π2-2θ                        Since, sin-1sinθ=θ, if θ-π2,π2   y=π2-2sin-1x            Since, x=sin θDifferentiating it with respect to x,  dydx=0-211-x2 dydx=-21-x2

Page No 10.63:

Question 9:

Differentiate

cos-1 xx2+a2

Answer:

Let,   y=cos-1xx2+a2Put   x=a cotθ     y=cos-1a cotθa2cot2θ+a2      y=cos-1a cotθa2cot2θ+1     y=sin-1a cotθa cosecθ     y=cos-1 cosθsinθ1sinθ     y=cos-1cosθ        y=θ     y=cot-1xa                  since, x=a cotθ   Differentiating it with respect to x using chain rule,dydx=-11+xa2ddxxadydx=-a2a2+x2×1adydx=-aa2+x2

Page No 10.63:

Question 10:

Differentiate

sin-1 sin x+cos x2,-3 π4<x<π4

Answer:

Let,     y=sin-1sinx+cosx2       y=sin-1sinx12+cosx12        y=sin-1sinx cosπ4+cosx sinπ4        y=sin-1sinx+π4    ...iHere, -3π4<x<π4   -3π4+π4<x+π4<π4+π4   -π2<x+π4<π2From i we get,  y= x+π4                       Since, sin-1sinθ=θ, if θ-π2,π2Differentiating it with respect to x,dydx=1+0dydx=1

Page No 10.63:

Question 11:

Differentiate

cos-1 cos x+sin x2, -π4<x<π4

Answer:

Let,     y=cos-1cosx+sinx2           y=cos-112cosx+12sinx            y=cos-1cosπ4cosx +sinπ4sinx             y=cos-1cosπ4-x            ...iHere, -π4<x<π4      π4>-x>-π4        -π4<-x<π4         -π4+π4<-x+π4< π4+π4         0<π4-x<π2So, from equation i,      y=  π4-x                      Since, cos-1cosθ=θ, if θ0, π Differentiating it with respect to x,dydx=0-1dydx=-1

Page No 10.63:

Question 12:

Differentiate

(i) tan-1 x1+1-x2, -1<x<1

(ii) tan-1 1+cosxsin x

Answer:

(i)
Let,   y=tan-1x1+1-x2Put   x=sinθ         y=tan-1sinθ1+1-sin2θ         y=tan-1 sinθ1+cosθ              y=tan-12 sinθ2cosθ22cos2θ2          y=tan-1tan θ2                ...iHere,  -1<x<1       -1<sinθ<1       -π2< θ<π2       -π4< θ2<π4So, from equation i,       y=θ2                        Since, tan-1tanθ=θ, if θ-π2,π2      y=12sin-1x            Since, x= sinθDifferentiating it with respect to x,dydx=121-x2

(ii)
Let y=tan-11+cosxsinx. Then,y=tan-12cos2x22sinx2.cosx2y=tan-1cotx2y=tan-1tanπ2-x2y=π2-x2dydx=0-12=-12

Page No 10.63:

Question 13:

Differentiate

tan-1 xa+a2-x2,-a<x<a

Answer:

Let,   y=tan-1xa+a2-x2Put   x=a sinθ     y=tan-1a sinθa+a2-a2sin2θ     y=tan-1a sinθa+a21-sin2θ          y=tan-1 a sinθa+a cosθ      y=tan-1a sinθa1+cosθ       y=tan-1 sinθ1+cosθ       y=tan-1 2sinθ2cosθ22 cos2θ2     y=tan-1tan θ2             ...i                   Here,  -a<x<a       -1<xa<1       -1< sinθ<1       -π2< θ<π2       -π4< θ2<π4So, from equation i,       y=θ2                        Since, tan-1tanθ=θ, if θ-π2,π2  y=12sin-1xa            Since, x=a sinθDifferentiating it with respect to x,  dydx=12×11-xa2ddxxadydx=a2a2-x2×1a dydx=12a2-x2

Page No 10.63:

Question 14:

Differentiate

sin-1 x+1-x22,-1<x<1

Answer:

Let, y=sin-1x+1-x22putting x=sinθ y=sin-1sinθ+1-sin2θ2 y=sin-1sinθ+cosθ2 y=sin-1sinθ12+cosθ12y=sin-1sinθ cosπ4+cosθ sinπ4y=sin-1sinθ+π4                  .....1Here, -1<x<1 -1<sinθ<1   -π2<θ<π2        -π2+π4<  π4+θ <3π4         -π4<  π4+θ <3π4So, from 1,      y=θ+π4                      Since, sin-1sinα=α, if α-π2,π2     y=sin-1x+π4          Differentiating it with respect to x, dydx=11-x2+0dydx=11-x2

Page No 10.63:

Question 15:

Differentiate

cos-1 x+1-x22, -1<x<1

Answer:

Let     y=cos-1x+1-x22Put,     x=cosθ          y=cos-1cosθ+1-cos2θ2           y=cos-1cosθ+sinθ2           y=cos-1cosθ12+sinθ12            y=cos-1cosθcosπ4+sinθ sinπ4            y=cos-1cosθ-π4                  ...iHere, -1<x<1      -1<cosθ<1        3π4<θ<5π4                3π4-π4<θ-π4<5π4-π4         π2< θ-π4<πSo, from equation i,      y=θ-π4                      Since, cos-1cosθ=θ, if θ0,π        y=cos-1x-π4             Since, x=sinθDifferentiating it with respect to x,dydx=-11-x2+0dydx=-11-x2

Page No 10.63:

Question 16:

Differentiate

tan-14x1-4 x2,-12<x<12

Answer:

Let,   y=tan-14x1-4x2put  2x=tanθ         y=tan-12 tanθ1-tan2θ        y=tan-1tan2θ           ...iHere,  -12<x<12       -1<2x<1        -1< tanθ<1        -π4< θ<π4        -π2< 2θ<π2So, from equation i,       y=2θ                        Since, tan-1tanθ=θ, if θ-π2,π2      y=2tan-12x            Since, 2x=tanθDifferentiating it with respect to x,dydx=211+2x2ddx2xdydx=211+4x2×2 dydx=41+4x2

Page No 10.63:

Question 17:

Differentiate

tan-1 2x+11-4x,-<x<0

Answer:

Let,   y=tan-12x+11-4xput   2x=tanθ         y=tan-12x×21-2x2         y=tan-12 tanθ1-tan2θ             y=tan-1tan2θ                   ...iHere,  -<x<0       2-<2x<2°        0< 2x<1        0< θ<π4        0<2θ<π2So, from equation i,       y=2θ                        Since, tan-1tanθ=θ, if θ-π2,π2      y=2 tan-12x           Differentiating it with respect to x,dydx=21+2x2ddx2xdydx=2×2x loge21+4xdydx=2x+1 loge21+4x

Page No 10.63:

Question 18:

Differentiate

tan-1 2 ax1-a2x, a>1, -<x<0

Answer:

Let,   y=tan-12ax1-a2xput   ax=tanθ         y=tan-12×ax1-ax2       y=tan-12 tanθ1-tan2θ            y=tan-1tan2θ                   ...iHere,  -<x<0       a-<ax<2°        0< tanθ<1        0< θ<π4        0<2θ<π2So, from equation i,       y=2θ                        Since, tan-1tanθ=θ, if θ-π2,π2      y=2 tan-1ax           Differentiating it with respect to x,dydx=21+ax2ddxaxdydx=2×ax logea1+a2xdydx=2ax logea1+a2x

Page No 10.63:

Question 19:

Differentiate

sin-1 1+x+1-x2,0<x<1

Answer:

Let,   y=sin-11+x+1-x2put    x=cos 2θ         y=sin-11+cos 2θ+1-cos 2θ2      y=sin-12cos2θ+2sin2θ2        y=sin-12 cosθ+2 sinθ2           y=sin-1cosθ12+12sinθ        y=sin-1 cosθ sinθπ4+cosπ4sinθ         y=sin-1sinθ+π4                   ...iHere, 0<x<1      0<cos 2θ<1        0< 2θ<π2        0< θ<π4     π4< θ+π4<π2So, from equation i,        y=θ+π4              Since, sin-1sinθ=θ, if θ-π2,π2     y=12cos-1x+π4Differentiate it with respect to x,dydx=12-11-x2+0dydx=-121-x2

Page No 10.63:

Question 20:

Differentiate

tan-1 1+a2 x2-1ax, x0

Answer:

Let,      y=tan-11+a2x2-1axput     ax=tanθ      y=tan-11+tan2θ-1tanθ       y=tan-1secθ-1tanθ       y=tan-11-cosθsinθ       y=tan-12sin2θ22θ2θ2      y=tan-1tanθ2      y=θ2      y=12tan-1ax  

Differentiate it with respect to x using chain rule,

dydx=12×11+ax2ddxaxdydx=121+a2x2adydx=a21+a2x2

Page No 10.63:

Question 21:

Differentiate

tan-1 sin x1+cos x, -π<x<π

Answer:

Let  fx=tan-1sinx1+cosx

This function is defined for all real numbers where cos x ≠ 1

fx=tan-1sinx1+cosxfx=tan-12 sinx2cosx22 cos2x2fx=tan-1tanx2=x2Thus, f'x=ddxx2=12

Page No 10.63:

Question 22:

Differentiate

sin-1 11+x2

Answer:

Let,  y=sin-111+x2put  x=cot θ     y=sin-111+cot2θ      y=sin-11cosec2θ     y=sin-1sinθ     y=θ    y=cot-1x            since, cotθ=x

Differentiate it with respect to x,

dydx=-11+x2

Page No 10.63:

Question 23:

Differentiate

cos-1 1-x2n1+x2n,<x<

Answer:

Let,     y=cos-11-x2n1+x2nPut    xn=tanθ      y=cos-11-xn21+xn2     y=cos-11-tan2θ1+tan2θ     y=cos-1cos 2θ         ...iHere,  0<x<       0<xn<       0<tanθ<       0<θ<π2       0<2θ<πSo, from equation i,    y=2θ                      Since, cos-1cos θ=θ, if θ0, π  y=2 tan-1xn

Differentiate it with respect to x using chain rule,

dydx=211+xn2ddxxndydx=21+x2n×nxn-1dydx=2nxn-11+x2n

Page No 10.63:

Question 24:

Differentiate

sin-1 1-x21+x2+sec-1 1+x21-x2, xR

Answer:

Let,    y=sin-11-x21+x2+sec-11+x21-x2    y=sin-11-x21+x2+cos-11-x21+x2         Since, sec-1x=cos-11x    y=π2                               Since, sin-1x+cos-1x=π2

Differentiate it with respect to x,

 dydx=0

Page No 10.63:

Question 25:

Differentiate

tan-1 a+x1-ax

Answer:

Let,     y=tan-1a+x1-ax       y=tan-1a+tan-1x                 Since, tan-1x+tan-1y=tan-1x+y1-xy

Differentiate it with respect to x,

 dydx=ddxtan-1a+ddxtan-1xdydx=0+11+x2 dydx=11+x2

Page No 10.63:

Question 26:

Differentiate

tan-1 x+a1-xa

Answer:

Let,    y=tan-1x+a1-xa     y=tan-1x+tan-1a          Since, tan-1x+tan-1y=tan-1x+y1-xy

Differentiate it with respect to x using chain rule,

dydx=ddxtan-1x+ddxtan-1adydx=11+x2ddxx+0dydx=11+x12xdydx=12x1+x

Page No 10.63:

Question 27:

Differentiate

tan-1 a+b tan xb-a tan x

Answer:

Let,   y=tan-1a+b tanxb-a tanx     y=tan-1a+b tanxbb-a tanxb    y=tan-1ab+tanx1-abtanx    y=tan-1tantan-1ab+ tanx1-tantan-1ab×tanx    y=tan-1tantan-1ab+x    y=tan-1ab+x 

Differentiate it with respect to x,

  dydx=0+1 dydx=1

Page No 10.63:

Question 28:

Differentiate

tan-1 a+bxb-ax

Answer:

Let,   y=tan-1a+bxb-ax     y=tan-1a+bxbb-axb    y=tan-1ab+bxbbb-axb    y=tan-1ab+x1-abx   y=tan-1ab+tan-1x                   Since, tan-1x+tan-1y=tan-1x+y1-xy

Differentiate it with respect to x,

 dydx=0+11+x2dydx=11+x2

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Question 29:

Differentiate

tan-1 x-ax+a

Answer:

Let,      y=tan-1x-ax+a      y=tan-1x-axx+ax     y=tan-1xx-axxx+ax      y=tan-11-ax1+1×ax    y=tan-11-tan-1ax  

Differentiate it with respect to x,

dydx=0-11+ax2ddxax dydx=-x2x2+a2-ax2 dydx=aa2+x2

Page No 10.63:

Question 30:

Differentiate

tan-1 x1+6 x2

Answer:

Let,    y=tan-1x1+6x2    y=tan-13x-2x1+3x2x   y=tan-13x-tan-12x           Since, tan-1x-tan-1y=tan-1x-y1+xy 

Differentiate it with respect to x using chain rule,

dydx=11+3x2ddx3x-11+2x2ddx2xdydx=11+9x23-11+4x22dydx=31+9x2-21+4x2



Page No 10.64:

Question 31:

Differentiate

tan-1 5 x1-6 x2, -16<x<16

Answer:

Let,     y=tan-15x1-6x2      y=tan-13x+2x1-3x2x     y=tan-13x+tan-12x               Since, tan-1x+tan-1y=tan-1x+y1-xy 

Differentiate it with respect to x using chain rule,

dydx=11+3x2ddx3x+11+2x2ddx2xdydx=11+9x2×3+11+4x2×2dydx=31+9x2+21+4x2

Page No 10.64:

Question 32:

Differentiate

tan-1 cos x+sin xcos x-sin x, π4<x<π4

Answer:

Let,     y=tan-1cosx+sinxcosx-sinx        y=tan-1cosx+sinxcosxcosx-sinxcosx       y=tan-1cosxcosx+sinxcosxcosxcosx-sinxcosx        y=tan-11+tanx1-tanx       y=tan-1tanπ4+tanx1-tanπ4tanx      y=tan-1tanπ4+x      y=π4+x    

Differentiate it with respect to x,

  dydx=0+1dydx=1

Page No 10.64:

Question 33:

Differentiate

tan-1 x1/3+a1/31-a x1/3

Answer:

Let,      y=tan-1x13+a131-ax13       y=tan-1x13+tan-1a13                Since, tan-1x+tan-1y=tan-1x+y1-xy

Differentiate it with respect to x using chain rule,

  dydx=11+x132×ddxx13+0dydx=13× x13-11+x23dydx=13x231+x23

Page No 10.64:

Question 34:

Differentiate

sin-1 11+x2

Answer:

Let y=sin-1 11+x2Putting x=cot θ θ=cot-1xy=sin-1 11+cot θ2=sin-1 11+cot2 θ=sin-1 1cosec θ=sin-1 sin θ=θ y=cot-1xDiff w.r.t.to x, we getdydx=-11+x2

Page No 10.64:

Question 35:

If y=sin-1 2x1+x2+sec-1 1+x21-x2, 0 <x<1,  prove that dydx=41+x2

Answer:

Let,      y=sin-12x1+x2+sec-11+x21-x2     y=sin-12x1+x2+cos-11-x21+x2Put,    x=tanθ       y=sin-12 tanθ1+tan2θ+cos-11-tan2θ1+tan2θ         y=sin-1sin 2θ+cos-1cos 2θ                ...iHere,  0<x<1       0<tanθ<1       0<θ<π4       0<2θ<π2So, from equation i,  y=2θ+2θ                     Since, sin-1sinθ=θ, if θ-π2,π2      cos-1cosθ=θ, if θ0,πy=4θy=4tan-1x                    Since, x=tanθ

Differentiate it with respect to x,

 dydx=41+x2

Page No 10.64:

Question 36:

If y=sin-1 x1+x2+cos-1 11+x2, 0<x<, prove that dydx=21+x2.

Answer:

Let,      y=sin-1x1+x2+cos-111+x2Put    x=tanθ        y=sin-1 tanθ1+tan2θ+cos-111+tan2θ      y=sin-1sinθcosθsecθ+cos-11secθ      y=sin-1sinθcosθ1cosθ+cos-1cosθ        y=sin-1sin θ+cos-1cos θ              ...iHere,  0<x<       0<tanθ<       0<θ<π2So, from equation i,y=θ+θ                     Since, sin-1sinθ=θ, if θ-π2,π2      cos-1cosθ=θ, if θ0,πy=2θy=2tan-1x                    Since, x=tanθ

Differentiate it with respect to x,

 dydx=21+x2

Page No 10.64:

Question 37:

Differentiate the following with respect to x:

i cos-1 sin x

(ii) cot-11-x1+x

Answer:

i Let, fx=cos-1sinxfx=cos-1cosπ2-xfx=π2-xThus, f'x=ddxπ2-x= -1
(ii) 
cot-11-x1+x=ddxtan-11+x1-x=11+1+x1-x2×1-x+1+x1-x2=1-x21-x2+1+x2×21-x2=1-x21-2x+x2+1+2x+x2=1-x221+x2×21-x2=11+x2
 

Page No 10.64:

Question 38:

If y=cot-1 1+sin x+1-sin x1+sin x-1-sin x, show that dydx is independent of x.

Answer:

Let,   y=cot-11+sinx+1-sinx1+sinx-1-sinx         ...iWe have, 1+sinx+1-sinx1+sinx-1-sinx           =1+sinx+1-sinx21+sinx-1-sinx1+sinx+1-sinx           =1+sinx+1-sinx+21-sinx1+sinx1+sinx-1-sinx           =2+21-sin2x2sinx           =1+cosxsinx           =2cos2x22sinx2cosx2            =cotx2Therefore, equation i becomes     y=cot-1cotx2y=x2dydx=12Hence, dydx is independent of x.

Page No 10.64:

Question 39:

If y=tan-1 2x1-x2+sec-1 1+x21-x2, x>0, prove that dydx=41+x2.

Answer:

Here,     y=tan-12x1-x2+sec-11+x21-x2          y=tan-12x1-x2+cos-11-x21+x2 Put       x=tanθ         y=tan-12tanθ1-tan2θ+cos-11-tan2θ1+tan2θ         y=tan-1tan 2θ+cos-1cos 2θ        y=2θ+2θ            y=4θ        y=4 tan-1x                     using, x=tanθ

Differentiate it with respect to x,

 dydx=41+x2

Page No 10.64:

Question 40:

If y=sec-1 x+1x-1+sin-1 x-1x+1, x>0. Find dydx.

Answer:

Here,   y=sec-1x+1x-1+sin-1x-1x+1       y=cos-1x-1x+1+sin-1x-1x+1           Since, sec-1x=cos-11x       y=π2                                                     Since, cos-1α+sin-1α=π2

Differentiate it with respect to x,

 dydx=0

Page No 10.64:

Question 41:

If y=sin 2 tan-1 1-x1+x, find dydx.

Answer:

Here,      y=sin2 tan-11-x1+xPut        x=cos 2θWe have,  y=sin2 tan-11-cos 2θ1+cos 2θ                 =sin2 tan-12 sin2θ2 cos2θ                 =sin2 tan-1tan2θ                  =sin2 tan-1tanθ                 =sin2θ                 =sin2×12cos-1x                      Since, x=cos 2θ                 =sinsin-11-x2                 =1-x2

Differentiate it with respect to x using chain rule,

  dydx=121-x2ddx1-x2dydx=121-x2-2xdydx=-x1-x2

Page No 10.64:

Question 42:

If y=cos-1 2x+2 cos-1 1-4x2, 0<x<12, find dydx.

Answer:

Here,    y=cos-12x+2 cos-11-4x2Put      2x=cosθ         y=cos-1cos θ+2 cos-11-cos2θ         y=cos-1cos θ+2 cos-1sinθ       y=cos-1cos θ+2 cos-1cosπ2-θ          ...iHere,   0<x<12        0<2x<1        0<cosθ<1        0<θ<π2and        0>-θ>-π2      π2>π2-θ>0     0<π2-θ<π2So, from equation i,  y=θ+2π2-θ                 Since, cos-1cosθ=θ, if θ0,π y=+π-2θ y=π-θ y=π-cos-12x                  Since, 2x=cosθ

Differentiate it with respect to x using chain rule,

  dydx=0--11-2x2ddx2xdydx=11-4x22dydx=21-4x2

Page No 10.64:

Question 43:

If the derivative of tan−1 (a + bx) takes the value 1 at x = 0, prove that 1 + a2 = b.

Answer:

Here,    ddxtan-1a+bx=1 at x=0    11+a+bx2ddxa+bxx=0=1  11+a+bx2×bx=0=1b1+a+02=1 b=1+a21+a2 = b

Page No 10.64:

Question 44:

If y=cos-1 2x+2 cos-1 1-4 x2,-12<x<0, find dydx.

Answer:

Here,    y=cos-12x+2 cos-11-4x2Put      2x=cosθ           y=cos-1cos θ+2 cos-11-cos2θ          y=cos-1cos θ+2 cos-1sinθ          y=cos-1cos θ+2 cos-1cosπ2-θ           ...iNow,     -12<x<0          -1<2x<0          -1<cosθ<0          π2<θ<πAnd          -π2>-θ>-π          π2-π2>π2-θ>π2-π         0>π2-θ>-π2      -π2<π2-θ<0So, from equation i,y=θ+2-π2-θ            Since, cos-1cosθ=θ, if θ0,π cos-1cosθ=-θ, if θ-π,0y=θ-2×π2+2θy=-π+3θy=-π+3cos-12x             Since, 2x=cosθ

Differentiate it with respect to x using chain rule,

dydx=0+3-11-2x2ddx2xdydx=-31-4x2×2dydx=-61-4x2

Page No 10.64:

Question 45:

If y=tan-1 1+x-1-x1+x+1-x, find dydx.

Answer:

Here,     y=tan-11+x-1-x1+x+1-xPut       x=cos2θ           y=tan-11+cos2θ-1-cos2θ1+cos2θ+1-cos2θ          =tan-12 cos2θ-2 sin2θ2 cos2θ+2 sin2θ          =tan-12cosθ-sinθ2cosθ+sinθ          =tan-1cosθ-sinθcosθcosθ+sinθcosθ              Dividing numerator and denominator by cosθ          =tan-1cosθcosθ-sinθcosθcosθcosθ+sinθcosθ           =tan-11-tanθ1+tanθ

            =tan-1tanπ4-tanθ1+tanπ4×tanθ             =tan-1tanπ4-θ             =π4-θ            =π4-12cos-1x                 Using x=cos2θ

Differentiate it with respect to x,

dydx=0-12-11-x2dydx=121-x2

Page No 10.64:

Question 46:

If y=cos-1 2x-3 1-x213, find dydx.

Answer:

Let     y=cos-12x-31-x213Put,     x=cosθ          y=cos-12cosθ-31-cos2θ13           y=cos-12cosθ-3sinθ13           y=cos-1cosθ213+sinθ313Let cosϕ=213 and sinϕ=313            y=cos-1cosθcosϕ+sinθ sinϕ            y=cos-1cosθ-ϕ                  ...i      y=θ-ϕ                             y=cos-1x-ϕ             Differentiating it with respect to x,dydx=-11-x2+0dydx=-11-x2

Page No 10.64:

Question 47:

Differentiate sin-12x+1·3x1+36x with respect to x.

Answer:

We have,
y=sin-12x+1·3x1+36xy=sin-12·6x1+62xput 6x=tanθθ=tan-16xNow,y=sin-12tanθ1+tan2θy=sin-1sin2θy=2θy=2tan-16xdydx=2×16x2×6xlog6dydx=2log66x36x

Page No 10.64:

Question 48:

If y=sin-1 6x1-9x2, -132<x<132, then find dydx.

Answer:

We have,

y=sin-1 6x1-9x2, -132<x<132

So, dydx=ddxsin-16x1-9x2=ddxsin-16x1-9x2=11-6x1-9x22×ddx6x1-9x2=11-36x21-9x2×6xddx1-9x2+1-9x2ddx6x=11-36x2-324x4×6x×121-9x2ddx1-9x2+1-9x26=11-36x2-324x4×6x×121-9x2×-18x+61-9x2=11-36x2-324x4×-54x21-9x2+61-9x2=11-36x2-324x4×-54x2+61-9x21-9x2=-54x2+6-54x21-9x21-36x2-324x4=6-108x21-9x21-36x2-324x4



Page No 10.74:

Question 1:

Find dydx in each of the following cases:

xy=c2

Answer:

We have, xy=c2
Differentiating with respect to x, we get,
   ddxxy=ddxc2xdydx+yddxx=0           Using product rulexdydx+y=0xdydx=-ydydx=-yx

Page No 10.74:

Question 2:

Find dydx in each of the following cases:

y3-3xy2=x3+3x2 y

Answer:

We have, y3-3xy2=x3+3x2y
Differentiating with respect to x, we get,
ddxy3-ddx3xy2=ddxx3+ddx3x2y3y2dydx-3xddxy2+y2ddxx=3x2+3x2ddxy+yddxx2           Using product rule3y2dydx-3x2ydydx+y2=3x2+3x2dydx+y2x3y2dydx-6xydydx-3y2=3x2+3x2dydx+6xy3y2dydx-6xydydx-3x2dydx=3x2+6xy+3y23dydxy2-2xy-x2=3x2+2xy+y2dydx=3x+y23y2-2xy-x2dydx=x+y2y2-2xy-x2

Page No 10.74:

Question 3:

Find dydx in each of the following cases:

x2/3+y2/3=a2/3

Answer:

We have,  x23+y23=a23
Differentiating it with respect to x, we get,

     ddxx23+ddxy23=ddxa2323x23-1+23y23-1dydx=023x-13+23y-13dydx=023y-13dydx=-23x-13dydx=-23x-13×32y-13dydx=-x-13y-13dydx=-y13x13dydx=-yx13

Page No 10.74:

Question 4:

Find dydx in each of the following cases:

4x+3y=log 4x-3y

Answer:

We have, 4x+3y= log4x-3y
Differentiating with respect to x, we get,

   ddx4x+ddx3y=ddxlog4x-3y4+3dydx=14x-3yddx4x-3y       4+3dydx=14x-3y4-3dydx3dydx+34x-3ydydx=44x-3y-43dydx1+14x-3y=414x-3y-13dydx4x-3y+14x-3y=41-4x+3y4x-3ydydx=431-4x+3y4x-3y4x-3y4x-3y+1dydx=431-4x+3y4x-3y+1

Page No 10.74:

Question 5:

Find dydx in each of the following cases:

x2a2+y2b2=1

Answer:

We have, x2a2+y2b2=1
Differentiating with respect to x, we get,
   ddxx2a2+y2b2=ddx1ddxx2a2+ddxy2b2=01a22x+1b22ydydx=02yb2dydx=-2xa2dydx=-2xa2b22ydydx=-b2xa2y

Page No 10.74:

Question 6:

Find dydx in each of the following cases:

x5+y5= 5 xy

Answer:

We have, x5+y5=5xy
Differentiating with respect to x, we get,

   ddxx5+ddxy5=ddx5xy5x4+5y4dydx=5xdydx+yddxx5x4+5y4dydx=5xdydx+y15x4+5y4dydx=5xdydx+5y5y4dydx-5xdydx=5y-5x45dydxy4-x=5y-x4dydx=5y-x45y4-xdydx=y-x4y4-x

Page No 10.74:

Question 7:

Find dydx in each of the following cases:

x+y2=2axy

Answer:

We have, x+y2=2axy
Differentiating with respect to x, we get,
ddxx+y2=ddx2axy2x+yddxx+y=2axdydx+yddxx        2x+y1+dydx=2axdydx+y12x+y+2x+ydydx=2axdydx+2aydydx2x+y-2ax=2ay-2x+ydydx=2ay-x-y2x+y-axdydx=ay-x-yx+y-ax

Page No 10.74:

Question 8:

Find dydx in each of the following cases:

x2+y22=xy

Answer:

We have, x2+y2=xy
Differentiating with respect to x, we get,
ddxx2+y22=ddxxy2x2+y2ddxx2+y2=xdydx+yddxx     2x2+y22x+2ydydx=xdydx+y14xx2+y2+4yx2+y2dydx=xdydx+y4yx2+y2dydx-xdydx=y-4xx2+y2dydx4yx2+y2-x=y-4xx2+y2dydx=y-4xx2+y24yx2+y2-xdydx=4xx2+y2-yx-4yx2+y2

Page No 10.74:

Question 9:

Find dydx in each of the following cases:

tan-1 x2+y2=a

Answer:

We have, tan-1x2+y2=a
Differentiate with respect to x, we get,
     ddxtan-1x2+y2=ddxa11+x2+y22×ddxx2+y2=011+x2+y222x+2ydydx=02x+2ydydx=0x+ydydx=0dydx=-xy

Page No 10.74:

Question 10:

Find dydx in each of the following cases:

ex-y=log xy

Answer:

We have, ex-y=logxy
Differentiate with respect to x,
  ddxex-y=ddxlogxyex-yddxx-y=1xy×ddxxy          ex-y1-dydx=yxyddxx-xdydxy2      ex-y-ex-ydydx=1xyy1-xdydxex-y-ex-ydydx=1x-1ydydx1ydydx-ex-ydydx=1x-ex-ydydx1y-ex-y1=1x-ex-y1dydx1-yex-yy=1-xex-yxdydx=yx1-xex-y1-yex-ydydx=-y-xxex-y-1yex-y-1dydx=yxxex-y-1yex-y-1

Page No 10.74:

Question 11:

Find dydx in each of the following cases:

sin xy+cos x+y=1

Answer:

We have, sinxy+cosx+y=1
Differentiating with respect to x, we get,
   ddxsinxy+ddxcosx+y=ddx1cosxyddxxy-sinx+yddxx+y=0     cosxyxdydx+yddxx-sinx+y1+dydx=0cosxyxdydx+y1-sinx+y-sinx+ydydx=0xcosxydydx+ycosxy-sinx+y-sinx+ydydx=0xcosxy-sinx+ydydx=sinx+y-ycosxydydx=sinx+y-y cosxyxcosxy-sinx+y

Page No 10.74:

Question 12:

If 1-x2+1-y2=a x-y, prove that dydx=1-y21-x2

Answer:

We have,  1-x2+1-y2=ax-yLet x=sinA , y=sinB1-sin2A+1-sin2B=asinA-sinBcosA+cosB=asinA-sinB          a=cosA+cosBsinA-sinBa=2 cosA+B2cosA-B22 cosA+B2sinA-B2               sinA-sinB=2 cosA+B2sinA-B2cosA+cosB=2 cosA+B2cosA-B2a=cotA-B2cot-1a=A-B22cot-1a=A-B2cot-1a=sin-1x-sin-1y                        x=sinA,y=sinB
Differentiating with respect to x, we get,
   ddx2cot-1a=ddxsin-1x-ddxsin-1y0=11-x2-11-y2dydx11-y2dydx=11-x2dydx=1-y21-x2dydx=1-y21-x2



Page No 10.75:

Question 13:

If y 1-x2+x 1-y2=1, prove that dydx=-1-y21-x2

Answer:

We have,  y1-x2+x1-y2=1Let x=sinA , y=sinBsinB1-sin2A+sinA1-sin2B=1sinBcosA+sinAcosB=1                 sinx+y=sinx cosy+cosxsinysinA+B=1A+B=sin-11sin-1x+sin-1y=π2                             x=sinA, y=sinB
Differentiate with respect to x,
ddxsin-1x+ddxsin-1y=ddxπ211-x2+11-y2dydx=0dydx=-1-y21-x2

Page No 10.75:

Question 14:

If xy=1, prove that dydx+y2=0

Answer:

We have, xy=1  
Differentiating with respect to x, we get,
   ddxxy=ddx1xdydx+yddxx=0            Using product rulexdydx+y1=0dydx=-yx dydx=-y1y                        x=1ydydx=-y2dydx+y2=0    

Page No 10.75:

Question 15:

If xy2=1, prove that 2dydx+y3=0

Answer:

We have, xy2=1           ...i
Differentiating with respect to x, we get,

   ddxxy2=ddx1xddxy2+y2ddxx=0            x2ydydx+y21=02xydydx=-y2dydx=-y22xydydx=-y2xput x=1y2 from equation idydx=-y21y22dydx=-y32dydx+y3=0

Page No 10.75:

Question 16:

If x 1+y+y 1+x=0, prove that 1+x2dydx+1=0

Answer:

We have,  x1+y+y1+x=0     x1+y=-y1+xSquaring both sides,we get,  x1+y2=-y1+x2x21+y=y21+xx2+x2y=y2+y2xx2-y2=y2x-x2yx-yx+y=xyy-xx+y=-xyy+xy=-xy1+x=-xy=-x1+x
Differentiating with respect to x, we get,
dydx=-1+xddxx--xddxx+11+x2dydx=-1+x1+x11+x2dydx=-1-x+x1+x2dydx=-11+x21+x2dydx=-11+x2dydx+1=0
         

Page No 10.75:

Question 17:

If log x2+y2=tan-1 yx, prove that dydx=x+yx-y

Answer:

We have, logx2+y2=tan-1xylogx2+y212=tan-1yx12logx2+y2=tan-1yx
Differentiate with respect to x, we get,
 12ddxlogx2+y2=ddxtan-1yx 121x2+y2ddxx2+y2=11+yx2ddxyx121x2+y22x+2ydydx=x2x2+y2xdydx-yddxxx21x2+y2x+ydydx=x2x2+y2xdydx-yddxxx21x2+y2x+ydydx=x2x2+y2xdydx-y1x2 x+ydydx=xdydx-yydydx-xdydx=-y-xdydxy-x=-y+xdydx=-y+xy-xdydx=x+yx-y

Page No 10.75:

Question 18:

If sec x+yx-y=a prove that dydx=yx

Answer:

We have, secx+yx-y=a x+yx-y=sec-1a    
Differentiate with respect to x, we get,
  x-yddxx+y-x+yddxx-yx-y2=0 x-y 1+dydx-x+y 1-dydx=0 x-y+x-ydydx-x+y+x+ydydx=0dydxx-y+x+y=x+y-x+ydydx2x=2ydydx=yx

Page No 10.75:

Question 19:

If tan-1 x2-y2x2+y2=a, prove that dydx=xy1-tan a1+tan a

Answer:

We have, tan-1x2-y2x2+y2=a x2-y2x2+y2=tana x2-y2=tanax2+y2
Differentiating with respect to x,
 ddxx2-y2=tanaddxx2+y2  2x-2ydydx=tana2x+2ydydx  2x-2ydydx=2xtana+2ytanadydx2ytanadydx+2ydydx=2x-2xtana2y1+tanadydx=2x1-tanadydx=xy1-tana1+tana
Hence proved

Page No 10.75:

Question 20:

If xy log x+y=1, prove that dydx=-y x2y+x+yx xy2+x+y

Answer:

We have, xy logx+y=1
Differentiating it with respect to x,
 ddxxy logx+y=ddx1  xyddxlogx+y+x logx+ydydx+y logx+yddxx=0          using chain rule and product rulexy1x+yddxx+y+x logx+ydydx+y logx+y1=0xyx+y 1+dydx+x logx+ydydx+y logx+y=0xyx+ydydx+xyx+y+x1xydydx+y1xy=0         xy logx+y=1dydxxyx+y+1y=-1x+xyx+y dydxxy2+x+yx+yy=-x+y+x2yxx+y dydx=-x+y+x2yxx+yyx+yxy2+x+y dydx=-yxx+y+x2yx+y+xy2
Hence proved

Page No 10.75:

Question 21:

If y=x sin a+y, prove that dydx=sin2 a+ysin a+y-y cos a+y

Answer:

We have, y=x sina+y  
Differentiate with respect to y,
dydx=ddxx sina+ydydx=xddxsina+y+sina+yddxx        using product rule and chain ruledydx=x cosa+yddxa+y+sina+y1dydx1-x cosa+y=sina+ydydx=sina+y1-x cosa+ydydx=sina+y1-ysina+y cosa+yy=xsina+y  dydx=sin2a+ysin a+y-y cos a+y
Hence proved

Page No 10.75:

Question 22:

If x sin a+y+sin a cos a+y=0, prove that dydx=sin2 a+ysin a

Answer:

We have, xsina+y+sinacosa+y=0   
Differentiate with respect to x,
ddxx sina+y+ddxsin a cosa+y=0xddxsin a+y+sina+yddxx+sin addxcosa+y=0      x cosa+yddxa+y+sina+y1+sin a-sina+yddxa+y=0x cosa+ydydx+sina+y- sinasina+ydydx=0dydxx cosa+y-sina sina+y=-sina+ydydx-sin acos2a+ysina+y-sin a sina+y=-sina+y    x=-sinacosa+ysina+y-dydxsin a cos2a+y+sin a sin2a+ysina+y=-sina+ydydx=sina+ysina+ysinacos2a+y+sin2a+ydydx=sin2a+ysina            

Page No 10.75:

Question 23:

If y=x sin y, prove that dydx=sin y1-x cos y

Answer:

We have, y=x siny
Differentiating with respect to x, we get,
dydx=ddxx sinydydx=xddxsiny+sinyddxxdydx=xcosydydx+sinydydx1-xcosy=sinydydx=siny1-xcosy

Page No 10.75:

Question 24:

If y x2+1=log x2+1-x, show that x2+1 dydx+xy+1=0

Answer:

We have, yx2+1=logx2+1-x
Differentiating with respect to x, we get,
ddxyx2+1=ddxlogx2+1-x        using product rule and chain ruleyddxx2+1+x2+1dydx=1x2+1-x×ddxx2+1-xy2x2+1×ddxx2+1+x2+1dydx=1x2+1-x×12x2+1ddxx2+1-12xy2x2+1+x2+1dydx=1x2+1-x2x2x2+1-1x2+1dydx=1x2+1-xx-x2+1x2+1-xyx2+1x2+1dydx=-1x2+1-xyx2+1x2+1dydx=-1+xyx2+1x2+1dydx=-1+xyx2+1dydx+1+xy=0

Page No 10.75:

Question 25:

If sin xy+yx=x2-y2, find dydx

Answer:

​We have, sinxy+yx=x2-y2
Differentiating with respect to x, we get,
ddxsin xy+ddxyx=ddxx2-ddxy2cosxyddxxy+xdydx-yddxxx2=2x-2ydydx      cosxyxdydx+yddxx+xdydx-y1x2=2x-2ydydxcosxyxdydx+y1+1x2xdydx-y=2x-2ydydxx cosxydydx+y cosxy+1xdydx-yx2=2x-2ydydxdydxx cosxy+1x+2y=yx2-y cosxy+2xdydxx2cosxy+1+2xyx=1x2y-x2y cosxy+2x3dydx=2x3+y-x2y cosxyxx2 cosxy+1+2xy

Page No 10.75:

Question 26:

If tan x+y+tan x-y=1, find dydx

Answer:

​We have, tanx+y+tanx-y=1
Differentiating with respect to x, we get,
ddxtanx+y+ddxtanx-y=ddx1sec2x+yddxx+y+sec2x-yddxx-y=0    sec2x+y1+dydx+sec2x-y1-dydx=0sec2x+ydydx-sec2x-ydydx=-sec2x+y+sec2x-ydydxsec2x+y-sec2x-y=-sec2x+y+sec2x-ydydx=sec2x+y+sec2x-ysec2x-y-sec2x+y

Page No 10.75:

Question 27:

If ex+ey=ex+y, prove that dydx=-exey-1eyex-1 or dydx+ey-x=0

Answer:

ex+ey=ex+yex+eydydx=ex+y1+dydxex+eydydx=ex+y+ex+ydydxeydydx-ex+ydydx=ex+y-exdydxey-ex+y=ex+y-exdydx=ex+y-exey-ex+y=exey-1ey1-ex=-exey-1eyex-1

Page No 10.75:

Question 28:

If cos y=x cos a+y, with cos a±1, prove that dydx=cos2 a+ysin a

Answer:

We have, cos y=x cosa+yDifferentiating with respect to x, we get,ddxcosy=ddxxcosa+y-sinydydx=cosa+yddxx+xddx cosa+y        -sinydydx=cosa+y+x-sina+ydydxxsina+y-sin ydydx=cosa+y    cosycosa+ysina+y-sin ydydx=cosa+y    cosy=x cosa+yx=cosycosa+y  cosysina+y-sinycosa+ydydx=cos2a+ysina+y-ydydx=cos2a+ydydx=cos2a+ysina    

Page No 10.75:

Question 29:

If sin2 y+cos xy=k, find dydx at x=1, y=π4.

Answer:

We have,

sin2 y+cos xy=k

By differentiating both sides with respect to x, we get

ddxsin2 y+ddxcos xy=dkdx2sin yddxsin y+-sin xyddxxy=02sin y cosy-sin xyxdydx+ydxdx=02sin y cosy-xsin xydydx-ysin xy=0xsin xydydx=2sin y cosy-ysin xy

dydx=2sin y cosy-ysin xyxsin xydydxx=1, y=π4=2sin y cosy-ysin xyxsin xy=2sinπ4 cosπ4-π4sin1×π41×sin1×π4=2×12 ×12-π4×121×12=2×12-π4212=11-π4212=42-π4212=42-π4=2-π4

Page No 10.75:

Question 30:

If y=logcos x sin x logsin x cos x-1+sin-1 2x1+x2, find dydx at x=π4.

Answer:

We have, y=logcosx sinxlogsinx cosx-1+sin-12x1+x2 y=logcosx sinxlogcosx sinx+sin-12x1+x2        logab=logba-1y=log sinxlog cosx2+sin-12x1+x2                  logab=logbloga

Differentiating with respect to x,

dydx=ddxlog sinxlog cosx2+ddxsin-12x1+x2dydx=2log sinxlog cosxddxlog sinxlog cosx+11-2x1+x22×ddx2x1+x2 dydx=2log sinxlog cosxlog cosxddxlog sinx-log sinxddxlog cosxlog cosx2+1+x21+x4-2x21+x22-2x2x1+x22   dydx=2log sinxlog cosxlog cosx×1sinxddxsinx-log sinx×1cosxddxcosxlog cosx2+1+x21+x4-2x21+x22-2x2x1+x22 dydx=2log sinxlog cosxlog cosx×cosxsinx+log sinx × sinxcosxlog cosx2+1+x21-x222+2x2-4x21+x22dydx=2log sinxlog cosx3cotx log cosx+tanx log sinx+21+x2put  x=π4dydx=2log sinπ4log cosπ43 cotπ4 log cosπ4+tanπ4 log sinπ4+211+π42dydx=21log1221× log12+1× log12+21616+π2     dydx=2×2log12log122+3216+π2dydx=41log12+3216+π2dydx=41-12log2+ 3216+π2dydx=-8log2+3216+π2So, dydxx=π4=8416+π2-1log2

Page No 10.75:

Question 31:

If y+x+y-x=c, show that dydx=yx-y2x2-1

Answer:

Here,       y+x+y-x=c
Differentiating with respect to x,
ddxy+x+ddxy-x=ddxc12y+xddxy+x+12y-xddxy-x=0    12y+xdydx+1+12y-xdydx-1=0dydx12y+x+dydx12y-x=12y-x-12y+xdydx×121y+x+1y-x=12y+x-y-xy-xy+xdydxy-x+y+xy+xy-x=y+x-y-xy-xy+xdydx=y+x-y-xy+x+y-x×y+x-y-xy+x-y-x           rationalizing the denominatordydx=y+x+y-x-2y+xy-xy+x-y+xdydx=2y-2y2-x22xdydx=2y2x-2y2-x22xdydx=yx-y2-x2x2dydx=yx-y2x2-1



Page No 10.88:

Question 1:

Differentiate

x1/x

Answer:

Let y=x1x           ...i
Taking log on both sides,
log y=log x1x log y=1xlog x            log ab=blog a
Differentiating with respect to x,
1ydydx=1xddxlog x+log xddxx-1       Using product rule1ydydx=1x×1x+log x×-1x2       1ydydx=1x2-logxx21ydydx=1-logxx2
dydx=x1x1-log xx                    [From (i)]

Page No 10.88:

Question 2:

Differentiate

xsin x

Answer:

Let y=xsinx       ...i
Taking log on both sides,
log y=log xsinxlog y=sinx log x                 log ab=b loga
Differentiating with respect to x, we get,

1ydydx=sinxddxlog x+log xddxsinx              using product rule1ydydx=sinx1x+log xcosxdydx=ysinxx+log xcosx
,dydx=xsinxsinxx+log xcosx                        [From (i)]

Page No 10.88:

Question 3:

Differentiate

1+cos xx

Answer:

Let y=1+cosxx         ...i
Taking log on both sides,
log y=log1+cosxxlog y=x log1+cosx  
Differentiating with respect to x,
1ydydx=xddxlog1+cosx+log1+cosxddxx     1ydydx=x×11+cosxddx1+cosx+log1+cosx11ydydx=x1+cosx0-sin x+log1+cosx1ydydx=log1+cosx-x sinx1+cosxdydx=ylog1+cosx-x sinx1+cosxdydx=1+cosxxlog1+cosx-x sinx1+cosx              using equation i

Page No 10.88:

Question 4:

Differentiate

xcos-1 x

Answer:

Let y=xcos-1x                 ...i
Taking log both sides,
log y=log xcos-1xlog y=cos-1x log x                 
Differentiating with respect to x,
1ydydx=cos-1xddxlog x+log xddxcos-1x                   1ydydx=cos-1x1x+log x-11-x21ydydx=cos-1xx-log x1-x2dydx=ycos-1xx-log x1-x2dydx=xcos-1xcos-1xx-log x1-x2                     Using equation i

Page No 10.88:

Question 5:

Differentiate

log xx

Answer:

Let y=logxx           ...i
Taking log on both sides,
logy=loglogxxlogy=xloglogx                   
Differentiating with respect to x using chain rule,
1ydydx=xddxloglogx+loglogxddxx1ydydx=x1logxddxlogx+loglogx11ydydx=xlogx1x+loglogx1ydydx=1logx+loglogxdydx=y1logx+loglogxdydx=logxx1logx+loglogx                    using equation i

Page No 10.88:

Question 6:

Differentiate

log xcos x

Answer:

Let y=logxcosx        ...i
Taking log on both sides,
logy=loglogxcosxlogy=cosx loglogx              
Differentiating with respect to x,
1ydydx=cosxddxloglogx+loglogxddxcosx1ydydx=cosxlogxddxlogx+loglogx×-sinx1ydydx=cosxlogx×1x-sinx loglogxdydx=ycosxx logx-sinx loglogxdydx=logxcosxcosxx logx-sinx loglogx                     using equation i

Page No 10.88:

Question 7:

Differentiate

sin xcos x

Answer:

Let y=sinxcosx          ...iTaking log on both sides,logy=logsinxcosxlogy=cosx log sinx                                      Differentiating with respect to x,1ydydx=cosxddxlog sinx+log sinxddxcosx          1ydydx=cosx1sinxddxsinx+log sinx-sinx1ydydx=cosxsinxcosx-sinx log sinxdydx=ycosx cotx-sinx log sinxdydx=sinxcosxcosx cotx-sinx log sinx                    using equation i

Page No 10.88:

Question 8:

Differentiate

ex log x

Answer:

Let y=ex logx     Taking log on both sides,logy=xlogxlogelogy=x logx                                        Differentiating with respect to x,1ydydx=xddxlogx+logxddxx           1ydydx=x1x+logx11ydydx=1+logxdydx=y1+logxdydx=exlogx1+logx                    using equation idydx=elogxx1+logx dydx=xx1+logx      

Page No 10.88:

Question 9:

Differentiate

sin xlog x

Answer:

Let y=sinxlogx        ...iTaking log on both sides,logy=logsinxlogxlogy=logx log sinx                                    Differentiating with respect to x,1ydydx=logxddxlog sinx+log sinxddxlogx           1ydydx=logx1sinxddxsinx+log sinx1x1ydydx=logxsinxcosx+ log sinxxdydx=ylogx cotx+log sinxxdydx=sinxlogxlogx cotx+log sinxx                    using equation i

Page No 10.88:

Question 10:

Differentiate

10log sin x

Answer:

Let y=10log sinx          ...iTaking log on both sides,logy=log10log sinxlogy=log sinx log10                                Differentiating with respect to x,1ydydx=log10ddxlog sinx    1ydydx=log101sinxddxsinx1ydydx=log101sinxcosxdydx=ylog10×cotxdydx=10log sinx×log10 ×cotx                    using equation i

Page No 10.88:

Question 11:

Differentiate

log xlog x

Answer:

Let y=logxlogx       ....iTaking log on both sides,logy=loglogxlogxlogy=logx log logx                                      Differentiating  both side with respect to x,1ydydx=loglogxddxlogx+logx ddxloglogx           dydx=yloglogx1x+logx1logxddxlogxdydx=y1xloglogx+1xdydx=logxlogx1+loglogxx                    using equation i

Page No 10.88:

Question 12:

Differentiate

1010x

Answer:

Let y=1010x        ...iTaking log on both sides,logy=loge1010xlogy=10x loge10                                      Differentiating with respect to x,1ydydx=loge10×10xloge10    1ydydx=10x×loge102dydx=y10x×loge102              dydx=1010x×10x×loge102          using equation i

Page No 10.88:

Question 13:

Differentiate

sin xx

Answer:

Let y=sin xx          sin-1y=xx         ...iTaking log on both sides,logsin-1y=logxxlogsin-1y=x logx                                       Differentiating with respect to x,1sin-1ydydxsin-1y=xddxlogx+logxddxx          1sin-1y×11-y2dydx=x1x+logxdydx=sin-1y1-y21+logxdydx=sin-1sin xx1-sin xx21+logxdydx=sin-1sinxx1-sin xx2 1+logx                   dydx=xxcosxx1+logx                                  using equation i

Page No 10.88:

Question 14:

Differentiate

sin-1 xx

Answer:

Let y=sin-1xx        ...iTaking log on both sides,logy=logsin-1xxlogy=x logsin-1x                                    Differentiating with respect to x,1ydydx=xddxlog sin-1x+log sin-1xddxx         1ydydx=x1sin-1xddxsin-1x+log sin-1x1ydydx=xsin-1x11-x2+ log sin-1xdydx=ylog sin-1x+xsin-1x1-x2dydx=sin-1xxlog sin-1x+xsin-1x1-x2                    using equation i

Page No 10.88:

Question 15:

Differentiate

xsin-1 x

Answer:

Let y=xsin-1x       ...iTaking log on both sides,logy=logxsin-1xlogy=sin-1x logx                                     Differentiating with respect to x,1ydydx=sin-1xddxlogx+logxddxsin-1x           1ydydx=sin-1x1x+logx11-x2dydx=ysin-1xx+logx1-x2dydx=xsin-1xsin-1xx+logx1-x2                    using equation i

Page No 10.88:

Question 16:

Differentiate

tan x1/x

Answer:

Let y=tanx1x        ...iTaking log on both sides,logy=logtanx1xlogy=1xlogtanx                                   Differentiating with respect to x,1ydydx=1xddxlogtanx+logtanxddx1x         1ydydx=1x×1tanxddxtanx+logtanx-1x21ydydx=1xtanxsec2x-logtanxx2dydx=ysec2xxtanx-logtan xx2dydx=tanx1x sec2xxtanx-logtan xx2                 using equation i

Page No 10.88:

Question 17:

Differentiate

xtan-1 x

Answer:

Let y=xtan-1x        ...iTaking log on both sides,logy=log xtan-1xlogy=tan-1x logx                                 Differentiating with respect to x,1ydydx=tan-1xddxlogx+logxddxtan-1x           1ydydx=tan-1x1x+logx11+x2dydx=ytan-1xx+logx1+x2dydx=xtan-1xtan-1xx+logx1+x2                using equation i

Page No 10.88:

Question 18:

​Differentiate

(i) xx x

(ii) xsin x-cos x+x2-1x2+1

(iii) xx cos x+x2+1x2-1

(iv) x cos xx+x sin x1/x

(v) x+1xx+x1+1x

(vi)  esin x+tan xx

(vii)  cos xx+sin x1/x

(viii) xx2-3+x-3x2

 

Answer:

(i)
Let y=xxx               ...iTaking log on both sides,logy=logxxxlogy=logxx+logx12         logy=x logx+12logx      
Differentiating with respect to x,
1ydydx=xddxlogx+logxddxx+12ddxlogx         1ydydx=x1x+logx1+121x1ydydx=1+logx+12xdydx=y1+logx+12xdydx=xxx1+logx+12x                    using equation idydx=xx+122x+12x+logx


ii    Let y=xsinx-cosx+x2-1x2+1 y=elogxsinx-cosx+x2-1x2+1y=esinx-cosxlogx+x2-1x2+1
Differentiate it with respect to x using chain rule,
dydx=ddxesinx-cosxlogx+ddxx2-1x2+1      =esinx-cosxlogxddxsinx-cosxlogx+x2+1ddxx2-1-x2-1ddxx2+1x2+12      =elogxsinx-cosxsinx-cosxddxlogx+logxddxsinx-cosx+x2+12x-x2-12xx2+12      =xsinx-cosxsinx-cosx1x+logxsinx+cosx+2x3+2x-2x3+2xx2+12      =xsinx-cosxsinx-cosxx+sinx+cosxlogx+4xx2+12

iii Let y=xx cosx+x2+1x2-1Also, Let u=xx cosx  and v=x2+1x2-1  y=u+vdydx=dudx+dvdx             ...iNow, u=xx cosxlog u=logxx cosxlog u=x cosx logx
Differentiating both sides with respect to x,
1ududx=cosx logxddxx+xlogxddxcosx+x cosxddxlogxdudx=ucosx logx+x-sinxlogx+x cosx1xdudx=xx cosxcosx logx-x sinx logx+cosxdudx=xx cosxcosx1+logx-x sinx logx            ...2Again, v=x2+1x2-1log v=logx2+1-logx2-1
Differentiating both sides with respect to x,
  1vdvdx=2xx2+1-2xx2-1dvdx=v2xx2-1-2xx2+1x2+1x2-1dvdx=x2+1x2-1-4xx2+1x2-1dvdx=-4xx2-12                       ...3From i,ii and iii, we obtaindydx=xx cosxcosx1+logx-x sinx logx-4xx2-12 

 iv   Let y=x cosxx+x sinx1x Also, Let u=x cosxx  and v=x sinx1x y=u+vdydx=dudx+dvdx             ...iNow,  u=x cosxxlog u=logx cosxxlog u=x logx cosxlog u=xlogx+log cosxlog u=xlogx+xlog cosx
Differentiate both sides with respect to x,
1ududx=ddxx logx+ddxx log cosxdudx=ulogxddxx+xddxlogx+log cosxddxx+xddxlog cosxdudx=x cosxxlogx1+x1x+log cosx1+x1cosxddxcosxdudx=x cosxxlogx+1+log cosx+xcosx-sinxdudx=x cosxx1+logx+log cosx-x tanxdudx=x cosxx1-x tanx+logx+log cosxdudx=x cosxx1-x tanx+logx cosx                                ...iiAgain,v=x sinx1xlog v=logx sinx1xlog v=1xlogx sinxlog v=1xlogx+log sinxlog v=1xlogx+1xlog sinx
Differentiating both sides with respect to x,
   1vdvdx=ddx1xlogx+ddx1xlogsinx1vdvdx=logxddx1x+1xddxlogx+logsinxddx1x+1xddxlogsinx1vdvdx=logx-1x2+1x1x+logsinx-1x2+1x1sinxddxsinx1vdvdx=1x21-logx+-logsinxx2+1x sinxcosxdvdx=x sinx1x1-logxx2+-logsinx+x cotxx2dvdx=x sinx1x1-logx-logsinx+x cotxx2dvdx=x sinx1x1-logxsinx+x cotxx2                              ...iiiFrom i,ii and iii, we obtaindydx=x cosxx1-x tanx+logx cosx+x sinx1xx cotx+1-logx sinxx2

v Let y=x+1x x+x1+1xAlso, Let u=x+1x x  and v=x1+1x y=u +vdydx=dudx+dvdx            ...iThen,  u=x+1x x log u=logx+1x xlog u=x logx+1x
Differentiate both sides with respect to x,
   1ududx=logx+1xddxx+xddxlogx+1x1ududx= logx+1x+x1x+1xddxx+1xdudx=ulogx+1x+xx+1x×1-1x2dudx=x+1xxlogx+1x+x-1xx+1xdudx=x+1xxlogx+1x+x2-1x2+1dudx=x+1xxx2-1x2+1+logx+1xAgain,  v=x1+1xlog v=logx1+1xlog v=1+1xlog x
Differentiating both sides with respect to x,
   1vdvdx=logxddx1+1x+1+1xddxlogx 1vdvdx=-1x2logx+1+1x1x 1vdvdx=-logxx2+1x+1x2dvdx=v-logx+x+1x2dvdx=x1+1xx+1-logxx2                 ...iiiFrom i,ii and iii, we obtaindydx=x+1xxx2-1x2+1+logx+1x+x1+1xx+1-logxx2

vi Let y=esinx+tanxx y=esinx+elogtanxxy=esinx+exlogtanx               
Differentiating with respect to x,
dydx=ddxesinx+ddxexlogtanx       =esinxddxsinx+exlogtanxddxx logtanx             =esinxcosx+elogtanxxxddxlogtanx+logtanxddxx       =esinxcosx+tanxxxtanxsec2x+logtanx       =esinxcosx+tanxxxsecxcosecx+logtanx

viii Let y=cosxx+sinx1xy=elogcosxx+elogsinx1xy=exlogcosx+e1xlogsinx
Differentiating with respect to x,
dydx=ddxex logcosx+ddxe1xlog sinx       =ex logcosx×ddxx logcosx+e1xlog sinxddx1xlogsinx       =elogcosxx× xddxlogcosx+logcosx × ddxx+elogsinx1x×1xddxlogsinx+logsinxddx1x       =cosxxx1cosxddxcosx+logcosx1+sin1x1x×1sinx×ddxsinx+logsinx-1x2       =cosxxx1cosx-sinx+logcosx+sinx1x1x×1sinxcosx-1x2logsinx       =cosxxlogcosx-x tanx+sinx1xcotxx-1x2logsinx

viii Let y=xx2-3+x-3x2 Also, let u=xx2-3 and v=x-3x2 y=u+v
Differentiate  both sides with respect to x,
dydx=dudx+dvdx      ...iNow, u=xx2-3log u=logxx2-3 log u=x2-3 logx
Differentiating with respect to x,
  1ududx=logxddxx2-3+x2-3ddxlogx1ududx=logx2x+x2-31xdudx=xx2-3x2-3x+2x logxAlso, v=x-3x2log v=logx-3x2log v=x2logx-3
Differentiating both sides with respect to x,
  1vdvdx=logx-3ddxx2+x2ddxlogx-31vdvdx=logx-3 2x+x21x-3ddxx-3dvdx=v2x logx-3+x2x-3×1dvdx=x-3x2x2x-3+2xlogx-3
 Substituing the expressions ofdudxand dvdxin equation idydx=xx2-3x2-3x+2x logx+x-3x2x2x-3+2x logx-3



Page No 10.89:

Question 19:

Find dydx

y=ex+10x+xx

Answer:

We have, y=ex+10x+xx y=ex+10x+elogxx         y=ex+10x+exlogx 
Differentiating with respect to x,
dydx=ddxex+ddx10x+ddxexlogx     =ex+10x log10+exlogxddxx logx     =ex+10x log10+exlogxxddxlogx+logxddxx             =ex+10x log10+elogxxx1x+logx     =ex+10x log10+xx1+logx     =ex+10x log10+xxloge+logx                                 logee=1     =ex+10x log10+xxlogex                                logA+logB=logAB

Page No 10.89:

Question 20:

Find dydx

y=xn+nx+xx+nn

Answer:

We have, y=xn+nx+xx+nny=xn+nx+elogxx+nn              y=xn+nx+exlogx+nn     
Differentiate with respect to x,
dydx=ddxxn+ddxnx+ddxexlogx+ddxnn     =nxn-1+nx logn+elogxxxddxlogx+logxddxx     =nxn-1+nx logn+xxx1x+logx     =nxn-1+nx logn+xx1+logx     =nxn-1+nx logn+xxloge+logx                      logee=1 and logA+logB=logAB     =nxn-1+nx logn+xxlogex

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Question 21:

Find dydx

y=x2-13 2x-1x-3 4x-1

Answer:

We have, y=x2-132x-1x-34x-1          ...iy=x2-132x-1x-3124x-112
Taking log on both sides,
log y=logx2-132x-1x-3124x-112logy=logx2-13+log2x-1-logx-312-log4x-112logy=3 logx2-1+log2x-1-12logx-3-12log4x-1
Differentiating with respect to x using chain rule,
1ydydx=3ddxlogx2-1+ddxlog2x-1-12ddxlogx-3-12log4x-11ydydx=31x2-1ddxx2-1+12x-1ddx2x-1-121x-3ddxx-3-1214x-1ddx4x-11ydydx=31x2-12x+12x-12-121x-31-1214x-141ydydx=6xx2-1+22x-1-12x-3-24x-1dydx=y6xx2-1+22x-1-12x-3-24x-1dydx=x2-132x-1x-34x-16xx2-1+22x-1-12x-3-24x-1                     using equation i

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Question 22:

Find dydx

y=eax·sec x·log x1-2x

Answer:

We have, y=eaxsecx logx1-2x       ...iy=eaxsecx logx1-2x12
Taking log on both sides,
logy=logeax+logsecx+log logx-12log1-2x        logy=ax+logsecx+loglogx-12log1-2x         
Differentiating with respect to x using chain rule,
1ydydx=ddxax+ddxlog secx+ddxlog logx-12log1-2x1ydydx=a+1secxddxsecx+1logxddxlogx-1211-2xddx1-2x1ydydx=a+secx tanxsecx+1logx1x-1211-2x-2dydx=ya+tanx+1x logx+11-2xdydx=eaxsecx logx1-2xa+tanx+1x logx+11-2x                 Using equation i

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Question 23:

Find dydx

y=e3x sin 4x·2x

Answer:

We have, y=e3x× sin4x × 2x                 ...i
Taking log on both sides,
logy=loge3x+logsin4x+log2x           logy=3x loge+logsin4x+x log2     logy=3x+logsin4x+x log2
Differentiating with respect to x,
1ydydx=ddx3x+ddxlog sin4x+ddxx log21ydydx=3+1sin4xddxsin4x+log211ydydx=3+1sin4xcos4xddx4x+log21ydydx=3+cot4x4+log21ydydx=3+4cot4x+log2dydx=y3+4cot4x+log2dydx=e3xsin4x2x3+4cot4x+log2                   Using equation i

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Question 24:

Find dydx

y=sin x sin 2x sin 3x sin 4x

Answer:

We have, y=sinx sin2x sin3x sin4x              ...i
Taking log on both sides,
logy=logsinx sin2x sin3x sin4xlogy=logsinx+logsin2x+logsin3x+logsin4x
Differentiating with respect to x using chain rule,
1ydydx=ddxlogsinx+ddxlogsin2x+ddxlogsin3x+ddxlogsin4x1ydydx=1sinxddxsinx+1sin2xddxsin2x+1sin3xddxsin3x+1sin4xddxsin4x1ydydx=1sinxcosx+1sin2xcos2xddx2x+1sin3xcos3xddx3x+1sin4xcos4xddx4x1ydydx=cotx+cot2x2+cot3x3+cot4x4dydx=ycotx+2cot2x+3cot3x+4cot4xdydx=sinx sin2x sin3x sin4xcotx+2cot2x+3cot3x+4cot4x                     Using equation i

Page No 10.89:

Question 25:

Find dydx

y=xsin x+sin xx

Answer:

Let y=xsinx+sinxxAlso, let u=xsinx  and v=sinxx y=u+vdydx=dudx+dvdx             ...iNow, u=xsinxTaking log on both sides,logu=logxsinxlogu=sinx logxDifferentiating both sides with respect to x,1ududx=logxddxsinx +sinxddxlogx                 dudx=ucosx logx+sinx1xdudx=xsinxcosx logx+sinxx           ...iiAgain, v=sinxxTaking log on both sides,logv=logsinxxlogv=x logsinxDifferentiating both sides with respect to x,1vdvdx=logsinxddxx+xddxlogsinxdvdx=vlogsinx+x1sinxddxsinxdvdx=sinxxlog sinx+xsinxcosxdvdx=sinxxlog sinx+x cotx                   ..iiiFrom i,iiand iii, we obtaindydx=xsinxcosx logx+sinxx+sinxxlog sinx+x cotx  

Page No 10.89:

Question 26:

Find dydx

y=sin xcos x+cos xsin x

Answer:

We have, y=sinxcosx+cosxsinxy=elogsinxcosx+elogcosxsinxy=ecosx logsinx+esinx logcosxDifferentiating with respect to x,dydx=ddxecosx logsinx+ddxesinx logcosx     =ecosx logsinxddxcosx logsinx+esinx logcosxddxsinx logcosx              =elogsinxcosxcosxddxlogsinx+logsinxddxcosx+elogcosxsinxsinxddxlogcosx+logcosxddxsinx             =sinxcosxcosx1sinxddxsinx+logsinx×-sinx+cosxsinxsinx1cosxddxcosx+logcosx×cosx     =sinxcosxcotx cosx-sinx logsinx+cosxsinxtanx-sinx+cosx logcosx     =sinxcosxcotx cosx-sinx logsinx+cosxsinxcosx logcosx-sinx tanx

Page No 10.89:

Question 27:

Find dydx

y=tan xcot x+cot xtan x

Answer:

We have, y=tanxcotx+cotxtanx y=elogtanxcotx+elogcotxtanxy=ecotx logtanx+etanx logcotx
Differentiating with respect to x using chain rule and product rule,
dydx=ddxecotx logtanx+ddxetanx logcotx     =ecotx logtanxddxcotx logtanx+etanx logcotxddxtanx logcotx     =elogtanxcotxcotxddxlog tanx+log tanxddxcotx+elogcotxtanxtanxddxlog cotx+logcotxddxtanx           =tanxcotxcotx×1tanxddxtanx+log tanx-cosec2x+cotxtanxtanx×1cotxddxcotx+log cotxsec2x     =tanxcotxcosec2xsec2xsec2x-cosec2x log tanx+cotxtanxsec2xcosec2x-cosec2x+sec2x log cotx     =tanxcotxcosec2x-cosec2x log tanx+cotxtanxsec2x log cotx-sec2x     =tanxcotxcosec2x1-log tanx+cotxtanxsec2x log cotx-1

Page No 10.89:

Question 28:

Find dydx

y=sin xx+sin-1 x

Answer:

We have, y=sinxx+sin-1x y=elogsinxx+sin-1xy=exlog sinx+sin-1x             
Differentiating with respect to x using chain rule,
dydx=ddxex log sinx+ddxsin-1x        =ex log sinxddxx log sinx+11-x2ddxx        =elogsinxxxddxlog sinx+log sinxddxx+11-x×12x                =sinxxx1sinxddxsinx+log sinx+12x-x2        =sinxxxsinxcosx+log sinx+12x-x2        =sinxxxcotx+log sinx+12x-x2

Page No 10.89:

Question 29:

Find dydx

(i) y=xcos x+sin xtan x
(ii) y=xx+sin xx

Answer:

​i We have, y=xcosx+sinxtanxy=elogxcosx+elogsinxtanx              y=ecosx logx+etanx log sinx
Differentiating with respect to x using chain rule,
dydx=ddxecosx logx+ddxetanx log sinx       =ecosx logxddxcosx logx+etanx log sinxddxtanx log sinx       =elogxcosxcosxddxlogx+logxddxcosx+elogsinxtanxtanxddxlog sinx+log sinxddxtanx                   =xcosxcosx1x+logx-sinx+sinxtanxtanx1sinxddxsinx+log sinxsec2x       =xcosxcosxx-sinx logx+sinxtanxtanx1sinxcosx+sec2x log sinx       =xcosxcosxx-sinx logx+sinxtanx1+sec2x log sinx

ii  We have, y=xx+sinxxy=elogxx+elogsinxxy=ex logx+ex log sinx          
Differentiating with respect to x using chain rule and product rule,
dydx=ddxex logx+ddxex log sinx       =ex logxddxx logx+ex log sinxddxx log sinx       =exlogxxddxlogx+logxddxx+elogsinxxxddxlog sinx+log sinxddxx               =xxx1x+logx1+sinxxx1sinxddxsinx+log sinx       =xx1+ logx+sinxxx1sinxcosx+ log sinx       =xx1+ logx+sinxxx cotx+ log sinx

Page No 10.89:

Question 30:

Find dydx

y=tan xlog x+cos2 π4

Answer:

We have,  y=tanxlogx+cos2π4 y=elogtanxlogx+cos2π4y=elogx log tanx+cos2π4        
Differentiating with respect to x using chain rule,
dydx=ddxelogx log tanx+ddxcos2π4       =elogx log tanxddxlogx log tanx+0       =elogtanxlogxlogxddxlog tanx+log tanxddxlogx              =tanxlogxlogx1tanxddxtanx+log tanx1x       =tanxlogxlogx1tanxsec2x+log tanxx       =tanxlogxlogxsec2xtanx+log tanxx

Page No 10.89:

Question 31:

Find dydx

y=xx+x1/x

Answer:

We have,  y=xx+x1xy=elogxx+elogx1x y=ex logx+e1xlogx            
Differentiating with respect to x using chain rule,
dydx=ddxex logx+ddxe1xlogx       =ex logxddxx logx+e1xlogxddx1xlogx       =elogxxxddxlogx+logxddxx+elogx1x1xddxlogx+logxddx1x               =xxx1x+logx1+x1x1x1x+logx-1x2       =xx1+logx+x1x1x2-1x2logx       =xx1+logx+x1x1-logxx2

Page No 10.89:

Question 32:

Find dydx

y=xlog x+log xx

Answer:

Let  y=xlogx+logxxAlso, let  u=logxx  and v=xlogxy=v+udydx=dvdx+dudx         ...iNow, u=logxxlogu=loglogxxlogu=xloglogx
Differentiating both sides with respect to x,
   1ududx=loglogxddxx+xddxloglogxdudx=uoglogx+x1logxddxlogxdudx=logxxloglogx+xlogx×1xdudx=logxxloglogx+1logx               ...iiAlso,  v=xlogxlogv=logxlogxlogv=logx logx=logx2
Differentiating both sides with respect to x,
    1vdvdx=ddxlogx2 1vdvdx=2logxddxlogxdvdx=2vlogx1xdvdx=2xlogxlogxxdvdx=2xlogxlogxx           ...iiiFrom i,ii and iii, we obtaindydx=2xlogxlogxx+logxx loglogx+1logx

Page No 10.89:

Question 33:

If x13 y7=x+y20, prove that dydx=yx

Answer:

We have,  x13y7=x+y20
Taking log on both sides,
 logx13y7=logx+y2013logx+7logy=20logx+y               
Differentiating with respect to x using chain rule,
13ddxlogx+7ddxlogy=20ddxlogx+y13x+7ydydx=20x+yddxx+y13x+7ydydx=20x+y1+dydx7ydydx-20x+ydydx=20x+y-13xdydx7y-20x+y=20x+y-13xdydx7x+y-20yyx+y=20x-13x+yxx+ydydx=20x-13x-13yxx+yyx+y7x+7y-20ydydx=yx7x-13y7x-13ydydx=yx

Page No 10.89:

Question 34:

If x16 y9=x2+y17, prove that xdydx=2 y

Answer:

We have,  x16y9=x2+y17
Taking log on both sides,
 logx16y9=logx2+y1716logx+9logy=17logx2+y                    
Differentiating with respect to x using chain rule,
16ddxlogx+9ddxlogy=17ddxlogx2+y16x+9ydydx=17x2+yddxx2+y16x+9ydydx=17x2+y2x+dydx9ydydx-17x2+ydydx=34xx2+y-16xdydx9y-17x2+y=34xx2+y-16xdydx9x2+y-17yyx2+y=34x2-16x2+yxx2+ydydx9x2+9y-17yyx2+y=34x2-16x2-16yxx2+ydydx9x2-8yyx2+y=18x2-16yxx2+ydydx=yx29x2-8y9x2-8ydydx=2yxxdydx=2y

Page No 10.89:

Question 35:

If y=sin xx, prove that dydx=cos xx·xx1+log x

Answer:

Let  y=sinxx          ...iAlso, Let  u=xx        ...iiTaking log on both sides,logu=logxxlogu=xlogx
Differentiating both sides with respect to x,
   1ududx=ddxx logx   1ududx=xddxlogx+logxddxx   1ududx=x1x+logx1   1ududx=1+logxdudx=u1+logxdudx=xx1+logx                  ...iii       using equation iiNow, using equation ii in equation i,y=sinuDifferentiating with respect to x,dydx=ddxsinudydx=cosududxUsing equation ii and iii,dydx=cosxx× xx1+logx

Page No 10.89:

Question 36:

If xx+yx=1, prove that dydx=-xx 1+log x+yx·log yx·yx-1

Answer:

We have,  xx+yx=1elogxx+elogyx=1ex logx+ex logy=1           
Differentiating with respect to x using chain rule,
ddxexlogx+ddxex logy=ddx1ex logxddxx logx+ex logyddxx logy=0ex logxxddxlogx+logxddxx+elogyxxddxlogy+logyddxx=0xxx1x+logx1+yxx1ydydx+logy1=0xx1+logx+yxxydydx+logy=0yx×xydydx=-xx1+logx+yxlogyxyx-1dydx=-xx1+logx+yxlogydydx=-xx1+logx+yxlogyxyx-1

Page No 10.89:

Question 37:

If xy·yx=1, prove that dydx=-y y+x log yx y log x+x

Answer:

We have, xy×yx=1
Taking log on both sides,
logxy×yx=log1ylogx+x logy=log1                  
Differentiating with respect to x ,
ddxy logx+ddxx logx=ddxlog1yddxlogx+logxdydx+xddxlogy+logyddxx=0y1x+logxdydx+x1ydydx+logy1=0yx+logxdydx+xydydx+logy=0dydxlogx+xy=-logy+yxdydxy logx+xy=-x logy+yxdydx=-yxx logy+yy logx+x

Page No 10.89:

Question 38:

If xy+yx=x+yx+y, find dydx

Answer:

We have,  xy+yx=x+yx+yelogxy+elogyx=elogx+yx+yey logx+ex logy=ex+y logx+y                   
 
Differentiating with respect to x using chain rule and product rule,
ddxey logx+ddxex logy=ddxex+ylogx+yey logxyddxlogx+logxdydx+ex logyxddxlogy+logyddxx=ex+ylogx+yddxx+ylogx+yelogxyy1x+logxdydx+elogxxydydx+logy1=elogx+yx+yx+yddxlogx+y+logx+yddxx+yxyyx+logxdydx+yxxydydx+logy=x+yx+yx+y1x+yddxx+y+logx+y1+dydxxy×yx+xy logxdydx+yx×xydydx+yxlogy=x+yx+y1×1+dydx+logx+y1+dydxxy-1× y+xylogxdydx+yx-1× xdydx+yxlogy=x+yx+y+x+yx+ydydx+x+yx+ylogx+y+x+yx+ylogx+ydydxdydxxylogx+xyx-1-x+yx+y1+logx+y=x+yx+y1+logx+y-xy-1×y-yxlogydydx=x+yx+y1+logx+y-yxy-1-yxlogyxylogx+xyx-1-x+yx+y1+logx+y

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Question 39:

If xmyn=1, prove that dydx=-mynx

Answer:

We have, xmyn=1
Taking log on both side,
logxmyn=log1m logx+n logy=log1
Differentiating with respect to x,
dydxm logx+ddxn logy=ddxlog1mx+nydydx=0dydx=-mx×yndydx=-mynx

Page No 10.89:

Question 40:

If yx=ey-x, prove that dydx=1+log y2log y

Answer:

We have,  yx=ey-x
Taking log on both sides,
 logyx=logey-x              xlogy=y-xlogexlogy=y-x                   ...i

Differentiating with respect to x,
ddxx logy=ddxy-xxddxlogy+logyddxx=dydx-1x1ydydx+logy1=dydx-1dydxxy-1=-1-logydydxy1+logyy-1=-1+logy                               Using i              dydx1-1-logy1+logy=-1+logydydx=-1+logy2-logydydx=1+logy2logy

Page No 10.89:

Question 41:

If sin xy=cos yx, prove that dydx=log cos y-y cot xlog sin x+x tan y

Answer:

We have, sinxy=cosyx
Taking log on both sides,
logsinxy=logcosyx               y logsinx=x logcosy   
Differentiating with respect to x,
ddxy log sinx=ddxx logcosyyddxlog sinx+log sinxdydx=xdydxlog cosy+log cosyddxx           y1sinxddxsinx+log sinxdydx=xcosyddxcosy+logcosy1ysinxcosx+log sinxdydx=xcosy-sinydydx+log cosyy cotx+log sinxdydx=-x tanydydx+log cosydydxlog sinx+x tany=log cosy-y cotxdydx=log cosy-y cotxlog sinx+x tany

Page No 10.89:

Question 42:

If cos xy=tan yx, prove that dydx=log tan y+y tan xlog cos x-x sec y cosec y

Answer:

We have,  cos xy=tan yx
Taking log  on both sides,
logcosxy=logtanyxy log cosx=x log tany              
Differentiating it with respect to x using chain,
ddxy log cosx=ddxx log tanyyddxlog cosx+log cosxdydx=xddxlog tany+log tanyddxxy1cosxddxcosx+log cosxdydx=x1tanyddxtany+log tanyycosx-sinx+log cosxdydx=xtanysec2ydydx+log tany-ytanx+log cosxdydx=secy cosecy×xdydx+log tanydydxlog cosx-x secy cosecy=log tany+y tanxdydx=log tany+y tanxlog cosx-xsecy cosecy

Page No 10.89:

Question 43:

If ex+ey=ex+y, prove that dydx+ey-x=0

Answer:

We have, ex+ey=ex+y        ...(1)
Differentiating both sides using chain rule,
ddxex+ddxey=ddxex+yex+eydydx=ex+yddxx+yex+eydydx=ex+y1+dydxeydydx-ex+ydydx=ex+y-exdydx=ex+y-exey-ex+ydydx=ex+y-ex+y+eyex+y-ex-ex+y            Using eqn.1dydx=ey-exdydx=-ey-xdydx+ey-x=0



Page No 10.90:

Question 44:

If ey=yx, prove that dydx=log y2log y-1

Answer:

We have,  ey=yx
Taking log on both sides,
logey=logyxy loge=x logy              y=x logy              ...i
Differentiating with respect to x,
dydx=ddxx logydydx=xdydxlogy+logyddxx            dydx=xydydx+logy dydx1-xy=logydydxy-xy=logydydx=y logyy-xdydx=y logyy-ylogy                          Using equation idydx=y logylogyy logy-ydydx=y logy2ylogy-1dydx=logy2logy-1

Page No 10.90:

Question 45:

If ex+y-x=0, prove that dydx=1-xx

Answer:

We have,  ex+y-x=0ex+y=x         ...1
Differentiating with respect to x using chain rule,
ddxex+y=ddxxex+yddxx+y=1x1+dydx=1                                  Using equation i1+dydx=1xdydx=1x-1dydx=1-xx

Page No 10.90:

Question 46:

If y=x sin a+y, prove that dydx=sin2 a+ysin a+y-y cos a+y

Answer:

We have, y=x sina+y
Differentiating with respect to x using chain rule,
dydx=xddxsina+y+sina+yddxx       dydx=x cosa+ydydx+sina+y1-x cosa+ydydx=sina+ydydx=sina+y1-x cosa+ydydx=sina+y1-ysina+ycosa+y                  ysina+y=xdydx=sin2a+ysina+y-y cosa+y

Page No 10.90:

Question 47:

If x sin a+y+sin a cos a+y=0, prove that dydx=sin2 a+ysin a

Answer:

We have, x sina+y+sina cosa+y=0
Differentiating with respect to x using chain rule,
ddxx sina+y+sina cosa+y=0xddxsina+y+sina+yddxx+sinaddxcosa+y+cosa+yddxsin a=0x cosa+y0+dydx+sina+y+sina-sina+ydydx+0=0x cosa+y-sina sina+ydydx+sina+y=0dydx=-sina+yx cosa+y-sina sina+ydydx=-sina+y-sina cosa+ysina+ycosa+y-sina sina+y                   x=-sina cosa+ysina+ydydx=sin2a+ysinacos2a+y+sinasin2a+ydydx=sin2a+ysinacos2a+y+sin2a+ydydx=sin2a+ysina                       cos2a+y+sin2a+y=1

Page No 10.90:

Question 48:

If sin xy=x+y, prove that dydx=1-x+y y cot xx+y log sin x-1

Answer:

We have,  sinxy=x+y
Taking log on both the sides,
logsinxy=logx+yylogsinx=logx+y                        
Differentiating with respect to x using chain rule,
ddxy logsinx=ddxlogx+yyddxlog sinx+log sinxdydx=1x+yddxx+yysinxddxsinx+log sinxdydx=1x+y1+dydxycosxsinx+log sinxdydx=1x+y+1x+ydydxdydxlog sinx-1x+y=1x+y-y cotxdydxx+ylog sinx-1x+y=1-yx+y cotxx+ydydx=1-yx+ycotxx+ylogsinx-1

Page No 10.90:

Question 49:

If xy log x+y=1, prove that dydx=-y x2y+x+yx xy2+x+y

Answer:

We have,  xy logx+y=1          ...i
Differentiating with respect to x using chain rule,
dydxxy logx+y=ddx1xyddxlogx+y+x logx+ydydx+y logx+yddxx=0xyx+y1+dydx+xlogx+ydydx+y logx+y1=0xyx+y1+dydx+xlogx+ydydx+y logx+y=0xyx+ydydx+xyx+y+x1xydydx+y1xy=0                     Using equation idydxxyx+y+1y=-1x+xyx+ydydxxy2+x+yx+yy=-x+y+x2yxx+ydydx=-yxx+y+x2yx+y+xy2

Page No 10.90:

Question 50:

If y=x sin y, prove that dydx=yx 1-x cos y

Answer:

We have,  y=x siny             ...i
Differentiating with respect to x,
dydx=ddxx sinydydx=xddxsiny+sinyddxx             dydx=x cosydydx+siny1dydx-x cosydydx=sinydydx1-x cosy=sinydydx=siny1-x cosydydx=yx1-x cosy   siny=yx

Page No 10.90:

Question 51:

Find the derivative of the function f (x) given by
fx=1+x 1+x2 1+x4 1+x8 and hence find f' (1)

Answer:

We have,  fx=1+x1+x21+x41+x8Taking log on both sides,log fx=log1+x+log1+x2+log1+x4+log1+x8ddxlog fx=ddxlog1+x+log1+x2+log1+x4+log1+x81fxf'x=11+x+2x1+x2+4x31+x4+8x71+x8f'x=1+x1+x21+x41+x811+x+2x1+x2+4x31+x4+8x71+x8f'1=1+11+121+141+1811+1+211+12+4131+14+8171+18f'1=2×2×2×212+22+42+82f'1=2×2×2×2×121+2+4+8f'1=8×15=120

 

Page No 10.90:

Question 52:

If y=logx2+x+1x2-x+1+23tan-13 x1-x2, find dydx.

Answer:

We have, y=logx2+x+1x2-x+1+23tan-13x1-x2
Differentiating with respect to x using chain rule,
dydx=ddxlogx2+x+1x2-x+1+23ddxtan-13x1-x2dydx=1x2+x+1x2-x+1ddxx2+x+1x2-x+1+2311+3x1-x22ddx3x1-x2dydx=x2-x+1x2+x+1x2-x+1ddxx2+x+1-x2+x+1ddxx2-x+1x2-x+12+231-x221+x4-2x2+3x2 1-x2ddx3x-3xddx1-x21-x22dydx=1x2+x+1x2-x+12x+1-x2+x+12x-1x2-x+1+231-x221+x2+x41-x23-3x-2x1-x22dydx=2x3-2x2+2x+x2-x+1-2x3-2x2-2x+x2+x+1x4+2x2+1-x2+233-3x2+23x21+x2+x4dydx=-2x2+2x4+x2+1+23x2+131+x2+x4dydx=21-x2x4+x2+1+2x2+11+x2+x4dydx=21-x2+x2+11+x2+x4dydx=41+x2+x4

Page No 10.90:

Question 53:

If y=sin x-cos xsin x-cos x, π4<x<3π4, find dydx.

Answer:

We have, y=sinx-cosxsinx-cosx              ...(i)
Taking log on both sides,
logy=logsinx-cosxsinx-cosxlogy=sinx-cosx logsinx-cosx

1ydydx=logsinx-cosxddxsinx-cosx+sinx-cosxddxlogsinx-cosx                    using product rule1ydydx=logsinx-cosxcosx+sinx+sinx-cosxsinx-cosxddxsinx-cosx1ydydx=cosx+sinx logsinx-cosx+cosx+sinx1ydydx=cosx+sinx1+logsinx-cosxdydx=ycosx+sinx1+logsinx-cosxdydx=sinx-cosxsinx-cosxcosx+sinx1+logsinx-cosx                        using equation i

Page No 10.90:

Question 54:

If xy=ex-y, find dydx.

Answer:

We have, xy=ex-y
Taking log on both sides,
   logxy=logex-ylogx+logy=x-ylogelogx+logy=x-y×1logx+logy=x-y

ddxlogx+ddxlogy=ddxx-dydx1x+1ydydx=1-dydx1+1ydydx=1-1xy+1ydydx=x-1xdydx=yx-1xy+1

Page No 10.90:

Question 55:

If yx+xy+xx=ab, find dydx.

Answer:

Given that  yx+xy+xx=abPutting u=yx, v=xy  and w=xx , we get u+v+w=abdudx+dvdx+dwdx=0                                ...iNow, u=yx 
Taking log on both sides,
log u=x log y

1ududx=xddxlog y+log yddxx           using product rule1ududx=x1ydydx+log y×1dudx=uxydydx+log ydudx=yxxydydx+log y                    ...iiAlso, v=xy
Taking log on both sides,
log v=y log x

1vdvdx=yddxlogx+logxdydx1vdvdx=y1x+logxdydxdvdx=vyx+logxdydxdvdx=xyyx+logxdydx                      ...iiiAgain,   w=xx
Taking log on both sides,
log w=x log x

1wdwdx=xddxlog x+log xddxx1wdwdx=x1x+logx1dwdx=w1+log xdwdx=xx 1+log x         ...ivFrom i,ii,iiiand iv, we have     yxxydydx+log y+xyyx+log xdydx+xx1+log x=0x.yx-1+xy.log xdydx=-xx1+log x-y.xy-1-yxlog y  dydx=-yx log y+y.xy-1+xx 1+log xx.yx-1+xy log x

Page No 10.90:

Question 56:

If cosxy=cosyx, find dydx

Answer:

cosxy=cosyxTaking log on both sides we getylogcosx=xlogcosydydxlogcosx-ytanx=logcosy-dydxlogcosx+xtanydydx=logcosy+ytanxdydxlogcosx+xtany=logcosy+ytanxdydx=logcosy+ytanxlogcosx+xtany

Page No 10.90:

Question 57:

 If cosy=xcosa+y, where cosa±1, prove that dydx=cos2a+ysina

Answer:

cosy=xcosa+y-sinydydx=cosa+y-xsina+ydydx-sinydydx+xsina+ydydx=cosa+ydydxcosycosa+ysina+y-siny=cosa+y       x=cosycosa+ydydxcosysina+y-sinycosa+ycosa+y=cosa+ydydxsina+y-ycosa+y=cosa+ydydxsinacosa+y=cosa+ydydx=cos2a+ysina

Page No 10.90:

Question 58:

If x-yexx-y=a, prove that ydydx+x=2y

Answer:

x-yexx-y=aTaking log on both sides, we getlogx-y+xx-y=loga1-dydxx-y+x-y-x1-dydxx-y2=01-dydxx-y+xdydx-yx-y2=0x-y1-dydx+xdydx-yx-y2=0x-xdydx-y+ydydx+xdydx-yx-y2=0x-xdydx-y+ydydx+xdydx-y=0ydydx+x=2y

Page No 10.90:

Question 59:

If x=exy, prove that dydx=x-yxlogx

Answer:

x=exyTaking logarithm on both sides, we getlogx=xyylogx=xlogxdydx+yx=1logxdydx=1-yxlogxdydx=x-yxxlogxdydx=x-ydydx=x-yxlogx

Page No 10.90:

Question 60:

If y=xtanx+x2+12, find dydx

Answer:

 y=xtanx+x2+12Taking log on both sides, we getlogy=tanxlogx+12logx2+121ydydx=tanxx+sec2xlogx+12×2x2+1×x1ydydx=tanxx+sec2xlogx+xx2+1dydx=xtanx+x2+12tanxx+sec2xlogx+xx2+1dydx=xtanxtanxx+sec2xlogx+x2x2+2

Page No 10.90:

Question 61:

If y=1+α1x-α+βx1x-α1x-β+γx21x-α1x-β1x-γ, find dydx.

Answer:

y=1+α1x-α+βx1x-α1x-β+γx21x-α1x-β1x-γy=1+αx1-αx+βx1-αx1-βx+γx1-αx1-βx1-γTaking log on both sides, we getlogy=log1+logαx-log1-αx+logβx-log1-αx-log1-βx+logγx-log1-αx-log1-βx-log1-γxlogy=logαx-3log1-αx+logβx-2log1-βx+logγx-log1-γx1ydydx=1x+3α1-αx+1x+2β1-βx+1x+γ1-γx1ydydx=1x3+3α1x-α+2β1x-β+γ1x-γdydx=yxα1x-α+β1x-β+γ1x-γ

Page No 10.90:

Question 62:

Differentiate
If xyyx = ab, find dydx.

Answer:


xy-yx=ab

eylnx-exlny=ab          elnfx=fx

Differentiating both sides with respect to x, we get

ddxeylnx-ddxexlny=ddxab

eylnxy×ddxlnx+lnx×dydx-exlnyx×ddxlny+lny×ddxx=0

xyy×1x+lnxdydx-yxx×1ydydx+lny=0

xy-1y+xylnxdydx-xyx-1dydx-yxlny=0

xylnx-xyx-1dydx=yxlny-xy-1y

dydx=yyx-1lny-xy-1xxy-1lnx-yx-1



Page No 10.98:

Question 1:

If y=x+x+x+... to , prove that dydx=12 y-1

Answer:

We have,  y=x+x+x+... to  y=x+ySquaring both sides, we get,        y2=x+y 2y dydx=1+dydxdydx2y-1=1dydx=12y-1

Page No 10.98:

Question 2:

If y=cos x+cos x+cos x+... to , prove that dydx=sin x1-2 y

Answer:

We have, y=cosx+cosx+cosx+... to y=cosx+ySquaring both sides, we get, y2=cosx+y 2y dydx=-sinx+dydxdydx2y-1=-sinxdydx=-sinx2y-1dydx=sinx1-2y

Page No 10.98:

Question 3:

If y=log x+log x+log x+.. to , prove that 2 y-1 dydx=1x

Answer:

We have,y=logx+logx+logx+... to y=logx+ySquaring both sides,we get,y2=logx+y2y dydx=1x+dydxdydx2y-1=1x

Page No 10.98:

Question 4:

If y=tan x+tan x+tan x+.. to , prove that dydx=sec2 x2 y-1

Answer:

We have, y=tanx+tanx+tanx+... to y=tanx+ySquaring both sides, we get, y2=tanx+y2y dydx=sec2x+dydxdydx2y-1=sec2xdydx=sec2x2y-1

Page No 10.98:

Question 5:

If y=sin xsin xsin x... , prove that y2 cot x1-y log sin x

Answer:

We have, y=sinxsinxsinx....y=sinxy
Taking log on both sides,
log y=logsinxylog y=y logsinx       

1ydydx=yddxlogsinx+log sinxdydx1ydydx=y1sinxddxsinx+log sinxdydxdydx1y-log sinx=ysinxcosxdydx1-y log sinxy=y cotxdydx=y2cotx1-y log sinx

Page No 10.98:

Question 6:

If y=tan xtan xtan x... , prove that dydx=2 at x=π4

Answer:

We have, y=tanxtanxtanx... y=tanxy


Taking log on both sides,
log y=logtanxylog y=y log tanx

Differentiating with respect to x using chain rule ,
1ydydx=yddxlog tanx+log tandydx1ydydx=ytanxddxtanx+log tandydxdydx1y-log tanx=ytanxsec2xdydx=ytanxsec2x×y1-ylog tanxNow, dydxx=π4=y sec2π4tanπ4×y1-y log tanπ4dydxx=π4=y22211-y log tan 1dydxx=π4=2121-0       yπ4=tanπ4tanπ4tanπ4...=1                   dydxx=π4=2                                              



Page No 10.99:

Question 7:

If y=exex+xeex+exxe, prove that dydx=exex·xexexx+ex·log x
                                 +xeex·eex1x+ex·log x+exxe xxe·xe-1 x+e log x

Answer:

We have, y=exex+xeex+exxey=u+v+wdydx=dudx+dvdx+dwdx    ...iwhere  u=exex, v=xeexand w=exxeNow,u=exex                                ...ii
Taking log on both sides,
logu=logexexlogu=xexlogelogu=xex             ...iii     
Taking log on both sides,
log logu=logxexlog logu=ex logx
Differentiating with respect to x,
1loguddxlogu=exddxlogx+logxddxex1logu1ududx=exx+ex logxdudx=uloguexx+ex logxdudx=exex×xexexx+ex logx                    ...A                                                             Using equation ii and iiiNow, v=xeex                       ...iv
Taking log on both sides,
log v=log xeexlog v=eexlogx

1vdvdx=eexddxlogx+logxddxeex1vdvdx=eex1x+logxeexddxexdvdx=veex1x+logxeexexdvdx=xeex×eex1x+exlogx              ...B                                                                      Using equation 4Now, w=exxe                     ...v
Taking log on both sides,
logw=logexxelogw=xxelogelogw=xxe          ...vi
Taking log on both sides,
log logw=logxxelog logw=xelogx

1logwddxlogw=xeddxlogx+logxddxxe1logw1wdwdx=xe1x+logxexe-1dwdx=w logwxe-1+e logxxe-1dwdx=exxexxexe-11+e logx                         ----C                                                                              using equation v,viUsing equation A,B and C in equation i, we getdydx=exexxexexx+exlogx+xeex×eex1x+ex logx+exxexxexe-11+e logx   

Page No 10.99:

Question 8:

If y=cos xcos xcos x..., prove that dydx=-y2 tan x1-y log cos x

Answer:

We have, y=cosxcosxcosx....
y=cosxy
Taking log on both sides,
log y=logcosxylog y=y logcosx          
Differentiating with respect to x using chain rule,
1ydydx=yddxlog cosx+log cosxdydx1ydydx=y1cosxddxcosx+log cosxdydxdydx1y-log cosx=ycosx-sinxdydx1-y log cosxy=-y tanxdydx=-y2tanx1-y log cosx



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