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Differentiate the following w.r.t x: $\sin^mx.\cos ^nx$.

$\begin{array}{1 1} (A)\;\sin^mx.\cos^nx(-n\tan x+m\cot x) \\ (B)\;-\sin^mx.\cos^nx(-n\tan x+m\cot x) \\(C)\;\sin^mx.\cos^nx(n\tan x+m\cot x) \\ (D)\;\sin^mx.\cos^nx(-n\tan x-m\cot x)\end{array} $

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  • Chain rule : Suppose $f$ is a real value function which is composite of two functions $u$ and $v$.
  • (i.e) $f=uov$,then $\large\frac{df}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
  • Product rule : $\large\frac{d}{dx}$$(uv)=u \large\frac{d}{dx}$$(v)+v\large\frac{d}{dx}$$(u)$
Step 1:
This can be differentiated by applying the product rule.
Let $u=\sin^mx$
By applying chain rule
$u'=m\sin^{m-1}x.\cos x$
Step 2:
Let $v=\cos^nx$
$v'=n\cos^{n-1}x.(-\sin x)$
Therefore $\large\frac{dy}{dx}$$=\sin^mx.n\cos^{n-1}x(-\sin x)+\cos^nx.m\sin^{m-1}x.\cos x$
On simplifying we get,
$\qquad\qquad\;\;\;=\sin^mx.\cos^nx(-n\tan x+m\cot x)$
answered Jun 27, 2013 by sreemathi.v
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