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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} $

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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}}$
$\quad\;\;=\large\frac{-\sin(\tan\sqrt{x+1}\times \sec^2\sqrt{x+1}}{2\sqrt{x+1}}$
answered Jun 27, 2013 by sreemathi.v
 
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