# Find the derivative of the following w.r.t x:1)sin2x 2)sinmx cosnx

$\frac{d}{dx}\left({\mathrm{sin}}^{2}x\right)=2\mathrm{sin}x*\frac{d}{dx}\mathrm{sin}x\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}x*\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=\mathrm{sin}2x$

$\frac{d}{dx}\left[{\mathrm{sin}}^{m}x{\mathrm{cos}}^{n}x\right]\phantom{\rule{0ex}{0ex}}=\frac{d}{dx}\left({\mathrm{sin}}^{m}x\right){\mathrm{cos}}^{n}x+{\mathrm{sin}}^{m}x.\frac{d}{dx}\left({\mathrm{cos}}^{n}x\right)\phantom{\rule{0ex}{0ex}}=m{\mathrm{sin}}^{m-1}x\left(\frac{d}{dx}\mathrm{sin}x\right){\mathrm{cos}}^{n}x+{\mathrm{sin}}^{m}x.\left(n{\mathrm{cos}}^{n-1}x.\left(\frac{d}{dx}\mathrm{cos}x\right)\right)\phantom{\rule{0ex}{0ex}}=m{\mathrm{sin}}^{m-1}x.\mathrm{cos}x.{\mathrm{cos}}^{n}x+n.{\mathrm{sin}}^{m}x.{\mathrm{cos}}^{n-1}x.\left(-\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=m{\mathrm{sin}}^{m-1}x.{\mathrm{cos}}^{n+1}x-n{\mathrm{sin}}^{m+1}x.{\mathrm{cos}}^{n-1}x$

hope this helps you

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1) d/dx (sin2x) = 2sinx . d/dx (sinx)= -2sinxcosx

2) d/dx(sinmxcosnx)= cosnx. d/dx (sinmx) + sinmx . d/dx (cosnx)= mcosnx .sinm-1x (cosx) - nsinmxcosn-1x (sinx)

=mcosn+1xsinm-1x -- nsinm+1xcosn-1x = cosn-1xsinm-1x(mcos2x -- nsin2x)

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