Prove the following

Let

$y=2^{\cos ^{2} x}$

Taking log on both sides, we get

$\log y=\log 2^{\cos ^{2} x} \Rightarrow \log y=\cos ^{2} x \cdot \log 2$

Now,

Differentiating both sides w.r.t. $x$

$\frac{1}{y} \cdot \frac{d y}{d x}=\log 2 \cdot \frac{d}{d x} \cos ^{2} x$

$\frac{1}{y} \cdot \frac{d y}{d x}=\log 2\left[2 \cos x \cdot \frac{d}{d x} \cos x\right]$

$\frac{1}{y} \cdot \frac{d y}{d x}=\log 2[2 \cos x(-\sin x)]$

$\frac{1}{y} \cdot \frac{d y}{d x}=\log 2(-\sin 2 x)$

$\frac{d y}{d x}=-y \cdot \log 2 \sin 2 x$

Thus, $\frac{d y}{d x}=-2^{\cos ^{2} x}(\log 2 \sin 2 x)$