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Differentiate the following w.r.t x: $(\sin x)^{\large\cos x}$.

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Toolbox:
  • Product rule : $\large\frac{d}{dx}$$(uv)=u.\large\frac{d}{dx}$$(v)+v.\large\frac{d}{dx}$$(u)$
  • If $y=[f(x)]^g(x)$.Taking $\log$ on both sides,we have $\log y=g(x).\log[f(x)].$
Step 1:
$y=(\sin x)^{\large \cos x}$
Take $\log$ on both sides.
$\log x^{\large a}=a\log x$
$\log y=\cos x.\log(\sin x)$
Now differentiating on both sides,
Apply product rule,
$(uv)'=u'v+vu'$
Step 2:
Let $u=\cos x$
$\large\frac{du}{dx}$$=-\sin x$
$v=\log(\sin x)$
$\large\frac{dv}{dx}$$=\large\frac{1}{\sin x}$$\cos x$
$\quad\;\;=\cot x$
Hence $\large\frac{1}{y}\frac{dy}{dx}$$=\cot x.\tan x+\log(\sin x).(-\sin x)$
$\Rightarrow \large\frac{1}{y}\frac{dy}{dx}$$=\cos x.\tan x-\sin x\log(\sin x)$
Therefore $\large\frac{dy}{dx}$$=y[\cos x\cot x-\sin x\log(\sin x)]$
Substituting for $y$,
$\large\frac{dy}{dx}$$=(\sin x)^{\large\cos x}[\cos x\cot x-\sin x\log(\sin x)]$
answered Jun 27, 2013 by sreemathi.v
 
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