Differentiate the function $\cos(\tan \sqrt{x + 1})$ with respect to $x$.

$\frac{-\sin(\tan \sqrt{x + 1}) \sec^2(\sqrt{x + 1})}{2\sqrt{x + 1}}$

The correct answer is Option (1) → $\frac{-\sin(\tan \sqrt{x + 1}) \sec^2(\sqrt{x + 1})}{2\sqrt{x + 1}}$ ##

Let $y = \cos(\tan \sqrt{x + 1})$

On differentiating w.r.t. $x$, we get

$∴\frac{dy}{dx} = \frac{d}{dx} \cos(\tan \sqrt{x + 1}) = -\sin(\tan \sqrt{x + 1}) \cdot \frac{d}{dx} (\tan \sqrt{x + 1})$

$= -\sin(\tan \sqrt{x + 1}) \cdot \sec^2 \sqrt{x + 1} \cdot \frac{d}{dx} (x + 1)^{1/2} \quad \left[ ∵\frac{d}{dx} (\tan x) = \sec^2 x \right]$

$= -\sin(\tan \sqrt{x + 1}) \cdot (\sec \sqrt{x + 1})^2 \cdot \frac{1}{2}(x + 1)^{-1/2} \cdot \frac{d}{dx} (x + 1)$

$= \frac{-1}{2\sqrt{x + 1}} \cdot \sin(\tan \sqrt{x + 1}) \cdot \sec^2 (\sqrt{x + 1})$