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