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 Class 12 Commerce Math are prepared by experts and are 100% accurate.

#### Question 1:

Find $\frac{dy}{dx}$, when

#### Question 2:

Find $\frac{dy}{dx}$, when

#### Question 3:

Find $\frac{dy}{dx}$, when

#### Question 4:

Find $\frac{dy}{dx}$, when

$\therefore \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\left(2{e}^{\theta }\mathrm{cos}\theta \right)}{a\left(2{e}^{\theta }\mathrm{sin}\theta \right)}=\mathrm{cot}\theta$

#### Question 5:

Find $\frac{dy}{dx}$, when

#### Question 6:

Find $\frac{dy}{dx}$, when

#### Question 7:

Find $\frac{dy}{dx}$, when

#### Question 8:

Find $\frac{dy}{dx}$, when

Differentiating with respect to t,

Differentiating it with respect to t,

#### Question 9:

Find $\frac{dy}{dx}$, when

#### Question 10:

Find $\frac{dy}{dx}$, when

Differentiating it with respect to $\theta$,

Differentiating it with respect to $\theta$ using chain rule,

#### Question 11:

Find $\frac{dy}{dx}$, when

#### Question 12:

Find $\frac{dy}{dx}$, when

#### Question 13:

Find $\frac{dy}{dx}$, when

If , prove that

If prove that

If prove that

#### Question 17:

If , prove that $\frac{dy}{dx}=\frac{x}{y}$

#### Question 18:

If , prove that $\frac{dy}{dx}=1$

#### Question 19:

If , find $\frac{dy}{dx}$

If

If

If

If

#### Question 24:

If , show that at

If

#### Question 28:

Write the derivative of sinx with respect to cosx

#### Question 29:

If x = a (2θ –  sin 2θ) and y = a (1 – cos 2θ), find $\frac{dy}{dx}$ when $\mathrm{\theta }=\frac{\pi }{3}$.

Given values are:
$x=a\left(2\theta -\mathrm{sin}2\theta \right)\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}y=a\left(1-\mathrm{cos}2\theta \right)$

Applying parametric differentiation

$\frac{dx}{d\theta }$= 2a − 2acos2$\theta$

$\frac{dy}{d\theta }$= 0 + 2asin2$\theta$

$\frac{dy}{dx}$

Now putting the value of $\theta$$\frac{\mathrm{\pi }}{3}$

So, $\frac{dy}{dx}$ is $\frac{1}{\sqrt{3}}$ at $\mathrm{\theta }=\frac{\pi }{3}$.

#### Question 1:

Differentiate x2 with respect to x3

#### Question 2:

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

#### Question 3:

Differentiate (log x)x with respect to log x

Taking log on both sides,

#### Question 4:

Differentiate with respect to
(i)
(ii)

Differentiating it with respect to x,

Differentiating it with respect to x,

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 5:

Differentiate with respect to $\sqrt{1-4{x}^{2}}$, if

(i)

(ii)

(iii)

Differentiating it with respect to x,

Differentiate it with respect to x,

Differentiate it with respect to x,

#### Question 6:

Differentiate with respect to , if

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 7:

Differentiate with respect to , if
(i)

(ii)

Differentiating it with respect to x,

Differentiating it with respect to x,

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 8:

Differentiate with respect to .

Taking log on both sides,

Differentiating it with respect to x using chain rule,

Taking log on both sides,

Differentiating it with respect to x using chain rule,

#### Question 9:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 10:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 11:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 12:

Differentiate with respect to

differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 13:

Differentiate with respect to

differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 14:

Differentiate with  respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 15:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 16:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 17:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 18:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### Question 19:

Differentiate with respect to

Differentiating it with respect to x,

#### Question 20:

Differentiate with respect to

Differentiating it with respect to x,

Differentiating it with respect to x,

#### 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

(d) 1/2 e

#### Question 2:

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

(b)

(c)

(d) none of these

(c)

We have,

#### Question 3:

The derivative of the function
(a) (2/3)1/2
(b) (1/3)1/2
(c) 31/2
(d) 61/2

(a) (2/3)1/2

#### Question 4:

Differential coefficient of sec is
(a) $\frac{x}{1+{x}^{2}}$

(b)

(c) $\frac{1}{\sqrt{1+{x}^{2}}}$

(d) $\frac{x}{\sqrt{1+{x}^{2}}}$

(d) $\frac{x}{\sqrt{1+{x}^{2}}}$

This is the equation of differential equation which have coefficient $\frac{x}{\sqrt{1+{x}^{2}}}$.

If
(a) − 1/4
(b) − 1/2
(c) 1/4
(d) 1/2

(d) 1/2

If

(a)

(b)

(c)

(d)

(a)

If is
(a)

(b)

(c) not defined

(d)

(d)

#### Question 8:

Given

(a) $f\text{'}\left(\frac{1}{2}\right)=f\text{'}\left(-\frac{1}{2}\right)$

(b) $f\left(\frac{1}{2}\right)=-f\text{'}\left(-\frac{1}{2}\right)$

(c) $f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}\right)$

(d) $f\left(\frac{1}{2}\right)=f\text{'}\left(-\frac{1}{2}\right)$

If
(a)
(b)
(c)
(d)

(d)

#### Question 10:

If

(a) $-\frac{2}{1+{x}^{2}}$

(b) $\frac{2}{1+{x}^{2}}$

(c) $\frac{1}{2-{x}^{2}}$

(d) $\frac{2}{2-{x}^{2}}$

(a) $-\frac{2}{1+{x}^{2}}$

#### Question 11:

The derivative of

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

(a) does not exist

For the curve
(a) 1/2
(b) 1
(c) −1
(d) 2

(c) −1

If
(a) 2
(b) − 2
(c) 1
(d) − 1]

(d) − 1

#### Question 14:

Let
(a) 1/2
(b) x
(c) $\frac{1-{x}^{2}}{{x}^{2}-4}$
(d) 1

(d) 1

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{2}{1+{x}^{2}}×\frac{1+{x}^{2}}{2}=1$

(a) 1/2
(b) − 1/2
(c) 1
(d) − 1

(b) − 1/2

#### Question 16:

equals

(a) $\frac{{x}^{2}-1}{{x}^{2}-4}$

(b) 1

(c) $\frac{{x}^{2}+1}{{x}^{2}-4}$

(d) ${e}^{x}\frac{{x}^{2}-1}{{x}^{2}-4}$

(a) $\frac{{x}^{2}-1}{{x}^{2}-4}$

If

(a)

(b)

(c)

(d)

(d)

#### Question 18:

If

(a) $-\frac{y}{x}$

(b)

(c)

(d) none of these

(a) $-\frac{y}{x}$

If

(a)

(b)

(c)

(d)

(b)

#### Question 20:

The derivative of with respect to is
(a) 2

(b)

(c) $2/x$

(d) $1-{x}^{2}$

(a) 2

#### Question 21:

If is equal to
(a)
(b)
(c)
(d) none of these

#### Question 22:

If $f\left(x\right)=\left|{x}^{2}-9x+20\right|$, then f' (x) is equal to
(a)
(b)
(c)
(d) none of these

#### Question 23:

If $f\left(x\right)=\sqrt{{x}^{2}-10x+25}$, then the derivative of f (x) in the interval [0, 7] is
(a) 1
(b) −1
(c) 0
(d) none of these

(d) none of these

#### Question 24:

If , then for x > 10, g ' (x) is equal to
(a) 1
(b) −1
(c) 0
(d) none of these

(c) 0

#### Question 25:

If , the f' (x) is equal to
(a) 1
(b) 0
(c) ${x}^{l+m+n}$
(d) none of these

(b) 0
We have,

#### Question 26:

If $y=\frac{1}{1+{x}^{a-b}{+}^{c-b}}+\frac{1}{1+{x}^{b-c}+{x}^{a-c}}+\frac{1}{1+{x}^{b-a}+{x}^{c-a}}$, then $\frac{dy}{dx}$ is equal to
(a) 1
(b) ${\left(a+b+c\right)}^{{x}^{a+b+c-1}}$
(c) 0
(d) none of these

(c) 0

#### Question 27:

If , then $\frac{dy}{dx}$ is equal to
(a)

(b) $\frac{{y}^{2}}{{x}^{2}}\sqrt{\frac{1-{y}^{6}}{1+{x}^{6}}}$

(c) $\frac{{x}^{2}}{{y}^{2}}\sqrt{\frac{1-{x}^{6}}{1-{y}^{6}}}$

(d) none of these

(a)

$⇒\frac{1}{\sqrt{1-{x}^{6}}}×\frac{d}{dx}\left({x}^{3}\right)-\frac{1}{\sqrt{1-{y}^{6}}}×\frac{d}{dx}\left({y}^{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\sqrt{1-{x}^{6}}}×3{x}^{2}-\frac{1}{\sqrt{1-{y}^{6}}}×3{y}^{2}×\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{{x}^{2}}{{y}^{2}}\sqrt{\frac{1-{y}^{6}}{1-{x}^{6}}}$

#### Question 28:

If , then the value of is given by
(a) ∞
(b) 1
(c) 0
(d) $\frac{1}{2}$

(b) 1

#### Question 29:

If is equal to

(a) $\frac{{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$

(b) $\frac{y}{x}$

(c) $\frac{x}{y}$

(d) none of these

(b) $\frac{y}{x}$

$⇒\frac{\left({x}^{2}+{y}^{2}\right)\left(2x-2y\frac{dy}{dx}\right)-\left({x}^{2}-{y}^{2}\right)\left(2x+2y\frac{dy}{dx}\right)}{{\left({x}^{2}+{y}^{2}\right)}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒\frac{2{x}^{3}-2{x}^{2}y\frac{dy}{dx}+2x{y}^{2}-2{y}^{3}\frac{dy}{dx}-2{x}^{3}-2{x}^{2}y\frac{dy}{dx}+2x{y}^{2}+2{y}^{3}\frac{dy}{dx}}{{\left({x}^{2}+{y}^{2}\right)}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒-4{x}^{2}y\frac{dy}{dx}+4x{y}^{2}=0\phantom{\rule{0ex}{0ex}}⇒-4{x}^{2}y\frac{dy}{dx}=-4x{y}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{4x{y}^{2}}{4{x}^{2}y}\phantom{\rule{0ex}{0ex}}\therefore \frac{dy}{dx}=\frac{y}{x}$

#### Question 30:

If is equal to
(a)

(b)

(c)

(d) none of these

(a)

We have,

#### Question 31:

If

(a) $\frac{4{x}^{3}}{1-{x}^{4}}$

(b) $-\frac{4x}{1-{x}^{4}}$

(c) $\frac{1}{4-{x}^{4}}$

(d) $-\frac{4{x}^{3}}{1-{x}^{4}}$

(b) $-\frac{4x}{1-{x}^{4}}$

#### Question 32:

If is equal to
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) none of these

(c) 1

#### Question 1:

If y = x

We know

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore y=x\left|x\right|=\left\{\begin{array}{ll}{x}^{2},& x\ge 0\\ -{x}^{2},& x<0\end{array}\right\$

$⇒\frac{dy}{dx}=\left\{\begin{array}{ll}2x,& x\ge 0\\ -2x,& x<0\end{array}\right\$

For x < 0,

$\frac{dy}{dx}=-2x$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=-1}=-2×\left(-1\right)=2$

If y = x

#### Question 2:

If y = 2x

We know

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore y=2x+\left|x\right|=\left\{\begin{array}{ll}2x+x,& x\ge 0\\ 2x-x,& x<0\end{array}\right\$

$⇒y=2x+\left|x\right|=\left\{\begin{array}{ll}3x,& x\ge 0\\ x,& x<0\end{array}\right\$

For x ≥ 0,

y = 3x

$⇒\frac{dy}{dx}=3$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=1}=3$

For x < 0,

y = x

$⇒\frac{dy}{dx}=1$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=-1}=1$

Thus, if y = 2x + |x|, then ${\left(\frac{dy}{dx}\right)}_{x=-1}=1$ and ${\left(\frac{dy}{dx}\right)}_{x=1}=3$.

If y = 2x

#### Question 3:

If f(x) =

$f\left(x\right)=\left|{x}^{2}-x\right|=\left\{\begin{array}{ll}{x}^{2}-x,& {x}^{2}-x\ge 0\\ -\left({x}^{2}-x\right),& {x}^{2}-x<0\end{array}\right\$

$⇒f\left(x\right)=\left|{x}^{2}-x\right|=\left\{\begin{array}{ll}{x}^{2}-x,& x\left(x-1\right)\ge 0\\ x-{x}^{2},& x\left(x-1\right)<0\end{array}\right\$

For x ≥ 1,

$f\left(x\right)={x}^{2}-x$

$⇒f\text{'}\left(x\right)=2x-1$

$\therefore f\text{'}\left(2\right)=2×2-1=3$

If f(x) =

#### Question 4:

If y = sinxo and $\frac{dy}{dx}$ = k cos xo , then k = ________________.

$y=\mathrm{sin}x°$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\mathrm{sin}\left(\frac{\mathrm{\pi }}{180}x\right)$

$⇒\frac{dy}{dx}=\mathrm{cos}\left(\frac{\mathrm{\pi }}{180}x\right)×\frac{d}{dx}\left(\frac{\mathrm{\pi }}{180}x\right)$

$⇒\frac{dy}{dx}=\mathrm{cos}x°×\frac{\mathrm{\pi }}{180}$

$⇒\frac{dy}{dx}=\frac{\mathrm{\pi }}{180}\mathrm{cos}x°$

Comparing with $\frac{dy}{dx}=k\mathrm{cos}x°$, we get

$k=\frac{\mathrm{\pi }}{180}$

If y = sinxo and $\frac{dy}{dx}$ = k cosxo , then k = .

#### Question 5:

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

$f\left(x\right)={e}^{x}g\left(x\right)$

Differentiating both sides with respect to x, we get

$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left[{e}^{x}g\left(x\right)\right]$

$⇒f\text{'}\left(x\right)={e}^{x}×\frac{d}{dx}g\left(x\right)+g\left(x\right)×\frac{d}{dx}{e}^{x}$

$⇒f\text{'}\left(x\right)={e}^{x}×g\text{'}\left(x\right)+g\left(x\right)×{e}^{x}$

Putting x = 0, we get

$f\text{'}\left(0\right)={e}^{0}×g\text{'}\left(0\right)+g\left(0\right)×{e}^{0}$

$⇒f\text{'}\left(0\right)=1×1+2×1$                 [g(0) = 2, g'(0) = 1 and e0 = 1]

$⇒f\text{'}\left(0\right)=1+2=3$

Thus, the value of $f\text{'}\left(0\right)$ is 3.

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

#### Question 6:

If ​f(x) = 3$\left|x+2\right|$, then '(-3) = ___________________.

We know

$\left|x+2\right|=\left\{\begin{array}{ll}x+2,& x\ge -2\\ -\left(x+2\right),& x<-2\end{array}\right\$

$\therefore f\left(x\right)=3\left|x+2\right|=\left\{\begin{array}{ll}3\left(x+2\right),& x\ge -2\\ -3\left(x+2\right),& x<-2\end{array}\right\$

$⇒f\text{'}\left(x\right)=\left\{\begin{array}{ll}3,& x\ge -2\\ -3,& x<-2\end{array}\right\$

For x < −2, $f\text{'}\left(x\right)=-3$

$\therefore f\text{'}\left(-3\right)=-3$

If ​f(x) = 3$\left|x+2\right|$, then '(−3) = ____−3____.

#### Question 7:

If f(1) = 3, f'(2) = 1, then

Disclaimer: The solution is provided for the following question.

If f(1) = 3, f'(1) = 1, then

Solution:

$\frac{d}{dx}\left\{\mathrm{ln}f\left({e}^{x}+2x\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{1}{f\left({e}^{x}+2x\right)}×\frac{d}{dx}f\left({e}^{x}+2x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{f\left({e}^{x}+2x\right)}×f\text{'}\left({e}^{x}+2x\right)\frac{d}{dx}\left({e}^{x}+2x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{f\left({e}^{x}+2x\right)}×f\text{'}\left({e}^{x}+2x\right)×\left({e}^{x}+2\right)$

Putting x = 0, we get

${\left[\frac{d}{dx}\left\{\mathrm{ln}f\left({e}^{x}+2x\right)\right\}\right]}_{x=0}$

$=\frac{1}{f\left({e}^{0}+2×0\right)}×f\text{'}\left({e}^{0}+2×0\right)×\left({e}^{0}+2\right)$

= 1

If f(1) = 3, f'(1) = 1, then .

#### Question 8:

If f(x) = x $\left|x\right|$, then ​f '(x) = _________________.

We know

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore f\left(x\right)=x\left|x\right|=\left\{\begin{array}{ll}{x}^{2},& x\ge 0\\ -{x}^{2},& x<0\end{array}\right\$

$⇒f\text{'}\left(x\right)=\left\{\begin{array}{ll}2x,& x\ge 0\\ -2x,& x<0\end{array}\right\$

Thus, $f\text{'}\left(x\right)=2x$ when x ≥ 0 and $f\text{'}\left(x\right)=-2x$ when x < 0.

If f(x) = $x\left|x\right|$, then ​f '(x) = 2x when x ≥ 0 and −2x when x < 0.

#### Question 9:

​If f(x) = $\left|x-1\right|+\left|x-3\right|$, then f '(2) = ______________________.

We have

$\left|x-1\right|=\left\{\begin{array}{ll}x-1,& x\ge 1\\ -\left(x-1\right),& x<1\end{array}\right\$

$\left|x-3\right|=\left\{\begin{array}{ll}x-3,& x\ge 3\\ -\left(x-3\right),& x<3\end{array}\right\$

$\therefore f\left(x\right)=\left|x-1\right|+\left|x-3\right|=\left\{\begin{array}{ll}-\left(x-1\right)-\left(x-3\right),& x<1\\ \left(x-1\right)-\left(x-3\right),& 1\le x<3\\ \left(x-1\right)+\left(x-3\right),& x\ge 3\end{array}\right\$

$⇒f\left(x\right)=\left|x-1\right|+\left|x-3\right|=\left\{\begin{array}{ll}-2x+4,& x<1\\ 2,& 1\le x<3\\ 2x-4,& x\ge 3\end{array}\right\$

For 1 ≤ x < 3, f(x)  = 2

$f\text{'}\left(x\right)=0$, for 1 ≤ x < 3

$⇒f\text{'}\left(2\right)=0$

Thus, the value of f '(2) is 0.

​If f(x) = $\left|x-1\right|+\left|x-3\right|$, then f '(2) = ___0___.

#### Question 10:

​If f(x) =

For $\frac{\mathrm{\pi }}{4},

$\mathrm{cos}x<\mathrm{sin}x$

$⇒\mathrm{cos}x-\mathrm{sin}x<0$

$\therefore f\left(x\right)=\left|\mathrm{cos}x-\mathrm{sin}x\right|=-\left(\mathrm{cos}x-\mathrm{sin}x\right)$

$⇒f\left(x\right)=-\mathrm{cos}x+\mathrm{sin}x$

Differentiating both sides with respect to x, we get

$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(-\mathrm{cos}x\right)+\frac{d}{dx}\left(\mathrm{sin}x\right)$

$⇒f\text{'}\left(x\right)=-\left(-\mathrm{sin}x\right)+\mathrm{cos}x$

$⇒f\text{'}\left(x\right)=\mathrm{sin}x+\mathrm{cos}x$

$\therefore f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=\mathrm{sin}\frac{\mathrm{\pi }}{3}+\mathrm{cos}\frac{\mathrm{\pi }}{3}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$         $\left(\frac{\mathrm{\pi }}{4}<\frac{\mathrm{\pi }}{3}\le \frac{\mathrm{\pi }}{2}\right)$

​If f(x) =

#### Question 11:

​If f(x) =

For $0\le x<\frac{\mathrm{\pi }}{2}$,

cosx > 0

$\therefore f\left(x\right)=\left|\mathrm{cos}x\right|=\mathrm{cos}x$

$⇒f\text{'}\left(x\right)=-\mathrm{sin}x$

$\therefore f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=-\mathrm{sin}\frac{\mathrm{\pi }}{4}=-\frac{1}{\sqrt{2}}$

​If f(x) =

#### Question 12:

The derivative of x2 with respect to x3 is __________________.

Let u(x) = x2 and v(x) = x3.

$u\left(x\right)={x}^{2}$

$⇒\frac{du}{dx}=2x$

$v\left(x\right)={x}^{3}$

$⇒\frac{dv}{dx}=3{x}^{2}$

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

$⇒\frac{du}{dv}=\frac{2x}{3{x}^{2}}=\frac{2}{3x}$

Thus, the derivative of x2 with respect to x3 is $\frac{2}{3x}$.

The derivative of x2 with respect to x3 is .

#### Question 13:

For the curve

$\sqrt{x}+\sqrt{y}=1$         (Given)

Differentiating both sides with respect to x, we get

$\frac{d}{dx}\sqrt{x}+\frac{d}{dx}\sqrt{y}=\frac{d}{dx}\left(1\right)$

$⇒\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$⇒\frac{1}{2\sqrt{y}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}}$

$⇒\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}=-\sqrt{\frac{y}{x}}$

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)}=-\sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}$

$⇒{\left(\frac{dy}{dx}\right)}_{\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)}=-1$

Thus, the value of $\frac{dy}{dx}$ at $\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$ is −1.

For the curve

#### Question 14:

​If f(x) =

For $-\frac{\mathrm{\pi }}{2},

$\mathrm{sin}x<0$

$\therefore f\left(x\right)=\left|\mathrm{sin}x\right|=-\mathrm{sin}x$

$⇒f\text{'}\left(x\right)=-\mathrm{cos}x$

If f(x) =

#### Question 15:

​If f(x) =

For $0,

$\mathrm{sin}x<\mathrm{cos}x$

$⇒\mathrm{sin}x-\mathrm{cos}x<0$

$\therefore f\left(x\right)=\left|\mathrm{sin}x-\mathrm{cos}x\right|=-\left(\mathrm{sin}x-\mathrm{cos}x\right)$

$⇒f\left(x\right)=-\mathrm{sin}x+\mathrm{cos}x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=-\mathrm{cos}x-\mathrm{sin}x$

$\therefore f\text{'}\left(\frac{\mathrm{\pi }}{6}\right)=-\mathrm{cos}\frac{\mathrm{\pi }}{6}-\mathrm{sin}\frac{\mathrm{\pi }}{6}=-\frac{\sqrt{3}}{2}-\frac{1}{2}=-\frac{1}{2}\left(\sqrt{3}+1\right)$

If f(x) =

#### Question 16:

If y = tan xo, then ${\left(\frac{dy}{dx}\right)}_{x={45}^{o}}$ = ________________________.

$y=\mathrm{tan}x°$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\mathrm{tan}\left(\frac{\mathrm{\pi }}{180}x\right)\phantom{\rule{0ex}{0ex}}$

$⇒\frac{dy}{dx}={\mathrm{sec}}^{2}\left(\frac{\mathrm{\pi }}{180}x\right)×\frac{d}{dx}\left(\frac{\mathrm{\pi }}{180}x\right)\phantom{\rule{0ex}{0ex}}$

$⇒\frac{dy}{dx}=\frac{\mathrm{\pi }}{180}{\mathrm{sec}}^{2}\left(\frac{\mathrm{\pi }}{180}x\right)$

$⇒\frac{dy}{dx}=\frac{\mathrm{\pi }}{180}{\mathrm{sec}}^{2}x°$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=45°}=\frac{\mathrm{\pi }}{180}{\mathrm{sec}}^{2}45°$

$⇒{\left(\frac{dy}{dx}\right)}_{x=45°}=\frac{\mathrm{\pi }}{180}×{\left(\sqrt{2}\right)}^{2}=\frac{\mathrm{\pi }}{90}$

If y = tanxº, then ${\left(\frac{dy}{dx}\right)}_{x={45}^{o}}$ = .

#### Question 17:

If y = sin-1(ex) + cos-1(ex), then $\frac{dy}{dx}$ = ____________________.

$y={\mathrm{sin}}^{-1}\left({e}^{x}\right)+{\mathrm{cos}}^{-1}\left({e}^{x}\right)$

$⇒y=\frac{\mathrm{\pi }}{2}$          $\left({\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=\frac{\mathrm{\pi }}{2}\right)$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}\right)$

$⇒\frac{dy}{dx}=0$

If y = sin−1(ex) + cos−1(ex), then $\frac{dy}{dx}$ = ____0____.

#### Question 18:

If y = sin-1(3x-4x3), $\frac{1}{2}$< 1, then $\frac{dy}{dx}$ = ______________________.

y = sin−1(3x − 4x3)

Let x = sinθ.

$\therefore y={\mathrm{sin}}^{-1}\left(3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta \right)$

$⇒y={\mathrm{sin}}^{-1}\left(\mathrm{sin}3\theta \right)$

Now,

$\frac{1}{2}

$⇒\frac{1}{2}<\mathrm{sin}\theta <1$

$⇒\frac{\mathrm{\pi }}{6}<\theta <\frac{\mathrm{\pi }}{2}$

$⇒\frac{\mathrm{\pi }}{2}<3\theta <\frac{3\mathrm{\pi }}{2}$

$\therefore y={\mathrm{sin}}^{-1}\left(\mathrm{sin}3\theta \right)$

$⇒y={\mathrm{sin}}^{-1}\left[\mathrm{sin}\left(\mathrm{\pi }-3\theta \right)\right]$

$⇒y=\mathrm{\pi }-3{\mathrm{sin}}^{-1}x$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=0-3×\frac{1}{\sqrt{1-{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{3}{\sqrt{1-{x}^{2}}}$

If y = sin−1(3x − 4x3), $\frac{1}{2}$< 1, then $\frac{dy}{dx}$ = .

#### Question 19:

If y = sec-1  is equal to ___________________.

We know

${\mathrm{sec}}^{-1}a={\mathrm{cos}}^{-1}\frac{1}{a}$

$\therefore {\mathrm{sec}}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)={\mathrm{cos}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$      .....(1)

Now,

$y={\mathrm{sec}}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$

$⇒y={\mathrm{cos}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$        [Using (1)]

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=0$

Thus, if $y={\mathrm{sec}}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$, then $\frac{dy}{dx}=0$.

If $y={\mathrm{sec}}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$ , then $\frac{dy}{dx}$ is equal to ___0___.

#### Question 20:

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

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

$u\left(x\right)=\mathrm{cos}x$

$⇒\frac{du}{dx}=\frac{d}{dx}\mathrm{cos}x$

$⇒\frac{du}{dx}=-\mathrm{sin}x$     .....(1)

$v\left(x\right)=\mathrm{sin}x$

$⇒\frac{dv}{dx}=\frac{d}{dx}\mathrm{sin}x$

$⇒\frac{dv}{dx}=\mathrm{cos}x$        .....(2)

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

$⇒\frac{du}{dv}=\frac{-\mathrm{sin}x}{\mathrm{cos}x}$       [From (1) and (2)]

$⇒\frac{du}{dv}=-\mathrm{tan}x$

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

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

#### Question 21:

The derivative of log10x with respect to x is ___________________.

Let $y={\mathrm{log}}_{10}x$.

$y={\mathrm{log}}_{10}x$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{{\mathrm{log}}_{e}x}{{\mathrm{log}}_{e}10}\right)$

$⇒\frac{dy}{dx}=\frac{1}{{\mathrm{log}}_{e}10}×\frac{d}{dx}\left({\mathrm{log}}_{e}x\right)$

$⇒\frac{dy}{dx}=\frac{1}{{\mathrm{log}}_{e}10}×\frac{1}{x}$

$⇒\frac{dy}{dx}=\frac{1}{x{\mathrm{log}}_{e}10}$

Thus, the derivative of ${\mathrm{log}}_{10}x$ with respect to x is $\frac{1}{x{\mathrm{log}}_{e}10}$.

The derivative of log10x with respect to x is .

#### Question 22:

$\frac{d}{dx}f\left({x}^{3}\right)=f\text{'}\left({x}^{3}\right)×\frac{d}{dx}{x}^{3}$

$⇒\frac{d}{dx}f\left({x}^{3}\right)=f\text{'}\left({x}^{3}\right)×3{x}^{2}$     .....(1)

Now,

$\frac{d}{dx}f\left(x\right)=f\text{'}\left(x\right)=\frac{1}{1+{x}^{2}}$          (Given)

Replacing x by x3, we get

$f\text{'}\left({x}^{3}\right)=\frac{1}{1+{\left({x}^{3}\right)}^{2}}$

$⇒f\text{'}\left({x}^{3}\right)=\frac{1}{1+{x}^{6}}$         .....(2)

From (1) and (2), we get

$\frac{d}{dx}f\left({x}^{3}\right)=\frac{1}{1+{x}^{6}}×3{x}^{2}$

$⇒\frac{d}{dx}f\left({x}^{3}\right)=\frac{3{x}^{2}}{1+{x}^{6}}$

#### Question 23:

If y = cos (sin x2), then  is equal to ______________________.

$y=\mathrm{cos}\left(\mathrm{sin}{x}^{2}\right)$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\mathrm{cos}\left(\mathrm{sin}{x}^{2}\right)$

$⇒\frac{dy}{dx}=-\mathrm{sin}\left(\mathrm{sin}{x}^{2}\right)×\frac{d}{dx}\mathrm{sin}{x}^{2}$

$⇒\frac{dy}{dx}=-\mathrm{sin}\left(\mathrm{sin}{x}^{2}\right)×\mathrm{cos}{x}^{2}×\frac{d}{dx}{x}^{2}$

$⇒\frac{dy}{dx}=-\mathrm{sin}\left(\mathrm{sin}{x}^{2}\right)×\mathrm{cos}{x}^{2}×2x$

$⇒\frac{dy}{dx}=-2x\mathrm{cos}{x}^{2}\mathrm{sin}\left(\mathrm{sin}{x}^{2}\right)$

Putting $x=\sqrt{\frac{\mathrm{\pi }}{2}}$, we get

${\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2×\sqrt{\frac{\mathrm{\pi }}{2}}×\mathrm{cos}{\left(\sqrt{\frac{\mathrm{\pi }}{2}}\right)}^{2}\mathrm{sin}\left[\mathrm{sin}{\left(\sqrt{\frac{\mathrm{\pi }}{2}}\right)}^{2}\right]$

$⇒{\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2×\sqrt{\frac{\mathrm{\pi }}{2}}×\mathrm{cos}\frac{\mathrm{\pi }}{2}×\mathrm{sin}\left(\mathrm{sin}\frac{\mathrm{\pi }}{2}\right)$

$⇒{\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=-2×\sqrt{\frac{\mathrm{\pi }}{2}}×0×\mathrm{sin}1$

$⇒{\left(\frac{dy}{dx}\right)}_{x=\sqrt{\frac{\mathrm{\pi }}{2}}}=0$

Thus, $\frac{dy}{dx}$ at $x=\sqrt{\frac{\mathrm{\pi }}{2}}$ is 0.

If y = cos (sin x2), then  is equal to ___0___.

#### Question 24:

If y = log = ___________________.

For $y=\mathrm{log}\left|x\right|$ to be defined,

$\left|x\right|\ne 0$

$⇒x\ne 0$

Now,

$y=\mathrm{log}\left|x\right|=\left\{\begin{array}{ll}\mathrm{log}x,& x>0\\ \mathrm{log}\left(-x\right),& x<0\end{array}\right\$

$\therefore \frac{dy}{dx}=\left\{\begin{array}{ll}\frac{1}{x},& x>0\\ \frac{1}{\left(-x\right)}×\left(-1\right),& x<0\end{array}\right\$

$⇒\frac{dy}{dx}=\left\{\begin{array}{ll}\frac{1}{x},& x>0\\ \frac{1}{x},& x<0\end{array}\right\$

If y = log = .

#### Question 25:

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

$f\left(x\right)=a{x}^{2}+bx+c$

$\therefore f\text{'}\left(x\right)=\frac{d}{dx}\left(a{x}^{2}+bx+c\right)$

$⇒f\text{'}\left(x\right)=2ax+b$

$\therefore f\text{'}\left(1\right)+f\text{'}\left(4\right)-f\text{'}\left(5\right)$

$=\left(2a×1+b\right)+\left(2a×4+b\right)-\left(2a×5+b\right)$

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

$=b$

Thus, the value of $f\text{'}\left(1\right)+f\text{'}\left(4\right)-f\text{'}\left(5\right)$ is b.

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

#### Question 26:

If '(1) = 2 and g'$\left(\sqrt{2}\right)$ = 4, then the derivative of f(tan x) with respect of g(secx) at x$\frac{\mathrm{\pi }}{4}$ is equal to ______________.

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

$u\left(x\right)=f\left(\mathrm{tan}x\right)$

$⇒\frac{du}{dx}=\frac{d}{dx}f\left(\mathrm{tan}x\right)$

$⇒\frac{du}{dx}=f\text{'}\left(\mathrm{tan}x\right)\frac{d}{dx}\mathrm{tan}x$

$⇒\frac{du}{dx}=f\text{'}\left(\mathrm{tan}x\right)×{\mathrm{sec}}^{2}x$          .....(1)

$v\left(x\right)=g\left(\mathrm{sec}x\right)$

$⇒\frac{dv}{dx}=\frac{d}{dx}g\left(\mathrm{sec}x\right)$

$⇒\frac{dv}{dx}=g\text{'}\left(\mathrm{sec}x\right)\frac{d}{dx}\mathrm{sec}x$

$⇒\frac{dv}{dx}=g\text{'}\left(\mathrm{sec}x\right)×\mathrm{sec}x\mathrm{tan}x$           .....(2)

$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

$⇒\frac{du}{dv}=\frac{{\mathrm{sec}}^{2}x×f\text{'}\left(\mathrm{tan}x\right)}{\mathrm{sec}x\mathrm{tan}x×g\text{'}\left(\mathrm{sec}x\right)}$

$⇒\frac{du}{dv}=\frac{\mathrm{sec}x×f\text{'}\left(\mathrm{tan}x\right)}{\mathrm{tan}x×g\text{'}\left(\mathrm{sec}x\right)}$

Putting $x=\frac{\mathrm{\pi }}{4}$, we get

${\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{sec}\frac{\mathrm{\pi }}{4}×f\text{'}\left(\mathrm{tan}\frac{\mathrm{\pi }}{4}\right)}{\mathrm{tan}\frac{\mathrm{\pi }}{4}×g\text{'}\left(\mathrm{sec}\frac{\mathrm{\pi }}{4}\right)}$

$⇒{\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\sqrt{2}×f\text{'}\left(1\right)}{1×g\text{'}\left(\sqrt{2}\right)}$

$⇒{\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{\sqrt{2}×2}{1×4}$

$⇒{\left(\frac{du}{dv}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{1}{\sqrt{2}}$

Thus, the derivative of f(tan x) with respect of g(secx) at x = $\frac{\mathrm{\pi }}{4}$ is $\frac{1}{\sqrt{2}}$.

If '(1) = 2 and g'$\left(\sqrt{2}\right)$ = 4, then the derivative of f(tan x) with respect of g(secx) at x = $\frac{\mathrm{\pi }}{4}$ is equal to .

#### Question 1:

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

Differentiating with respect to x,

#### Question 2:

If $f\left(x\right)=x+1$, then write the value of .

$⇒\frac{d}{dx}\left\{\left(fof\right)\left(x\right)\right\}=\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\frac{d}{dx}\left\{\left(fof\right)\left(x\right)\right\}=1+0\phantom{\rule{0ex}{0ex}}⇒\frac{d}{dx}\left\{\left(fof\right)\left(x\right)\right\}=1$

#### Question 3:

If .

Differentiate it with respect to x,

#### Question 4:

If , find the value of the derivative of w.r. to x at the point x = 0.

If , then find .

#### 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).

#### Question 7:

If . Then, write the value of

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=1$

#### Question 8:

If

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(\mathrm{\pi }-\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=0-1\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-1$

#### Question 9:

If .

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(2\mathrm{\pi }-\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=0-1\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-1$

#### Question 10:

If , write the value of .

#### Question 11:

If , write the value of .

#### Question 12:

If , find .

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(-{x}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-2x$

#### Question 13:

If , find $\frac{dy}{dx}$.

$⇒\frac{dy}{dx}=0\phantom{\rule{0ex}{0ex}}$

If .

#### Question 15:

If .

$⇒\frac{dy}{dx}=-1$

#### Question 16:

If .

Taking log on both sides,

If .

If

If

#### Question 20:

If .

$⇒\frac{dy}{dx}=0$

#### Question 21:

If , then write the value of $\frac{dy}{dx}.$

$⇒\frac{dy}{dx}=0$

#### Question 22:

If to ∞, then find the value of $\frac{dy}{dx}$.

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{1-x}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{1}{{\left(1-x\right)}^{2}}\frac{d}{dx}\left(1-x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{1}{{\left(1-x\right)}^{2}}\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1}{{\left(1-x\right)}^{2}}$

#### Question 23:

If , where $-1, then write the value of $\frac{du}{dv}$.

#### Question 24:

If , then find the value of f' (1).

#### Question 25:

If

$⇒\frac{dy}{dx}=\frac{d}{dx}\left(\mathrm{log}\left|3x\right|\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1}{3x}\frac{d}{dx}\left(3x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1}{3x}\left(3\right)\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{1}{x}$

#### Question 26:

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

#### Question 27:

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

If

#### Question 29:

If y = log (cos ex), then find $\frac{dy}{dx}.$

$y=\mathrm{log}\left(\mathrm{cos}{e}^{x}\right)$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\mathrm{log}\left(\mathrm{cos}{e}^{x}\right)$

$⇒\frac{dy}{dx}=\frac{1}{\mathrm{cos}{e}^{x}}×\frac{d}{dx}\mathrm{cos}{e}^{x}$

$⇒\frac{dy}{dx}=\frac{1}{\mathrm{cos}{e}^{x}}×\left(-\mathrm{sin}{e}^{x}\right)×\frac{d}{dx}{e}^{x}$

$⇒\frac{dy}{dx}=-\frac{\mathrm{sin}{e}^{x}}{\mathrm{cos}{e}^{x}}×{e}^{x}$

$⇒\frac{dy}{dx}=-{e}^{x}\mathrm{tan}{e}^{x}$

#### Question 30:

If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find $\frac{d}{dx}\left(fog\right)\left(x\right).$

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

$fog\left(x\right)$

$=f\left(g\left(x\right)\right)$

$=x$

$\therefore \frac{d}{dx}\left(fog\right)\left(x\right)=\frac{d}{dx}\left(x\right)$

$⇒\frac{d}{dx}\left(fog\right)\left(x\right)=1$

Thus, the value of $\frac{d}{dx}\left(fog\right)\left(x\right)$ is 1.

#### Question 1:

Differentiate the following functions from first principles:

ex

#### Question 2:

Differentiate the following functions from first principles:

e3x

#### Question 3:

Differentiate the following functions from first principles:

eax+b

#### Question 4:

Differentiate the following functions from first principles:

ecos x

#### Question 5:

Differentiate the following functions from first principles:

${e}^{\sqrt{2x}}$

#### Question 6:

Differentiate the following functions from first principles:

log cos x

#### Question 7:

Differentiate the following function from first principles:

#### Question 8:

Differentiate the following functions from first principles:

x
2ex

#### Question 9:

Differentiate the following functions from first principles:

log cosec x

#### Question 10:

Differentiate the following functions from first principles:

sin−1 (2x + 3)

Differentiate

sin (3x + 5)

Differentiate

tan2 x

Differentiate

tan (x° + 45°)

Differentiate

sin (log x)

Differentiate

Differentiate

etan x

Differentiate

sin2 (2x + 1)

Differentiate

log7 (2x − 3)

Differentiate

tan 5x°

#### Question 10:

Differentiate

${{2}^{x}}^{3}$

#### Question 11:

Differentiate

${3}^{{e}^{x}}$

Differentiate

logx 3

#### Question 13:

Differentiate

${3}^{{x}^{2}+2x}$

#### Question 14:

Differentiate

$\sqrt{\frac{{a}^{2}-{x}^{2}}{{a}^{2}+{x}^{2}}}$

Differentiate

Differentiate

#### Question 17:

Differentiate

$\sqrt{\frac{1-{x}^{2}}{1+{x}^{2}}}$

Differentiate

(log sin x)2

#### Question 19:

Differentiate

$\sqrt{\frac{1+x}{1-x}}$

Differentiate

Differentiate

Differentiate

sin (log sin x)

Differentiate

Differentiate

Differentiate

Differentiate

Differentiate

Differentiate

Differentiate

Differentiate

#### Question 31:

Differentiate

$\frac{{e}^{2x}+{e}^{-2x}}{{e}^{2x}-{e}^{-2x}}$

Differentiate with respect to x we get,

#### Question 32:

Differentiate

Differentiate with respect of x we get,

#### Question 33:

Differentiate

Differentiate it with respect to x we get,

#### Question 34:

Differentiate

Differentiate it with respect to x we get,

#### Question 35:

Differentiate

Differentiate it with respect to x we get,

#### Question 36:

Differentiate

Differentiate it with respect to x we get,

#### Question 37:

Differentiate

$\sqrt{{\mathrm{tan}}^{-1}\left(\frac{x}{2}\right)}$

Differentiate it with respect to x we get,

Differentiate