# Differentiate the following w.r.t $x$ : $\cos\bigg(\tan\sqrt {x+1}\bigg)$.

$\begin{array}{1 1} (A)\;\large\frac{-\sin(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}} \\ (B)\;\large\frac{\sin(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}} \\ (C)\;\large\frac{-\cos(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}} \\ (D)\;\large\frac{\cos(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}} \end{array}$

Toolbox:
• Chain Rule : Suppose $f$ is a real valued function which is a composite of three functions $u,v$ and $w$ $(i.e$) $f=(won)ov$ then $\large\frac{df}{dx}=\frac{dw}{ds}$$\times \large\frac{ds}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
Given : $\cos\bigg(\tan\sqrt {x+1}\bigg)$.
Let $u=\sqrt{x+1}=(x+1)^{\large\frac{1}{2}}$
$\Rightarrow \large\frac{du}{dx}=\large\frac{1}{2}$$(x+1)^{-\large\frac{1}{2}} \qquad\quad=\large\frac{1}{2\sqrt{x+1}} Let v=\tan u \Rightarrow \large\frac{dv}{du}$$=\sec^2u$
Let $y=\cos v$
$\Rightarrow \large\frac{dy}{dv}$$=-\sin v Step 2: Therefore \large\frac{dy}{dx}=\frac{dy}{dv}$$\times \large\frac{dv}{du}$$\times \large\frac{du}{dx} \qquad\qquad\;\;\;\;=-\sin v\times \sec^2u\times \large\frac{1}{2\sqrt{x+1}} On substituting for v and u we get, \large\frac{dy}{dx}$$=-\sin(\tan x)\times \sec^2\sqrt{x+1}\times \large\frac{1}{2\sqrt{x+1}}$
$\quad\;\;=\large\frac{-\sin(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}}$