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Differentiate the following w.r.t $x$: $2^{\large \cos^2x}$

$\begin{array}{1 1} (A) \;2\cos^2x\log 2(\sin 2x) \\(B) \;-2^{\large\cos^2x}\log 2(\sin 2x) \\ (C)\;2\cos^2x\log 2(\sin 2x) \\ (D)\;-2\cos^2x\log 2(\cos 2x) \end{array} $

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  • $\large\frac{d}{dx}$$(\cos x)=-\sin x$
  • $2\sin x\cos x=\sin 2x$
Step 1:
Let us take $\log$ on both sides
$\log y=\cos^2x.\log 2$
$\log(x)^a=a\log x$
Now differentiating on both sides we get,
$\large\frac{1}{y}\frac{dy}{dx}=$$(2\cos x.\sin x)\log 2$
But $2\sin x\cos x=\sin 2x$
$\Rightarrow \large\frac{dy}{dx}$$=-y\log 2(\sin 2x)$
Step 2:
Substituting for $y$ we get,
$\large\frac{dy}{dx}$$=-2^{\large\cos^2 x}\log 2(\sin 2x)$
answered Jun 27, 2013 by sreemathi.v
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