Question:
$\cos (\tan \sqrt{x+1})$
Solution:
Let $y=\cos (\tan \sqrt{x+1})$
Differentiating both sides w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x} \cos (\tan \sqrt{x+1})$
$=-\sin (\tan \sqrt{x+1}) \cdot \frac{d}{d x}(\tan \sqrt{x+1})$
$=-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{d}{d x} \sqrt{x+1}$
$=-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{1}{2 \sqrt{x+1}} \cdot 1$
Thus, $\quad \frac{d y}{d x}=-\frac{1}{2 \sqrt{x+1}} \sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1}$
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