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Differentiate the following w.r.t $x$ : $\sin\sqrt x+\cos^2\sqrt x$.

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  • Chain Rule : Suppose a real valued function which is a composite of two function $u$ and $v$ $(i.e$) $f=vou$ ,then $\large\frac{df}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
$y=\sin\sqrt x+\cos^2\sqrt x$
Let $u=\sqrt x$
$\large\frac{dy}{du}=\frac{1}{2\sqrt x}$
$y=\sin u+\cos^2u$
$\large\frac{dy}{du}=$$\cos u+2\cos u(-\sin u)$
But $2\sin u\cos u=\sin 2u$
$\qquad=\cos u-\sin 2u$
Step 2:
$\large\frac{dy}{dx}=\frac{dy}{du}$$\times \large\frac{du}{dx}$
$\large\frac{dy}{dx}=$$(\cos u-\sin 2u)\times \large\frac{1}{2\sqrt x}$
Substituting for $u$ we get,
$\large\frac{dy}{dx}=\frac{\cos\sqrt x-\sin 2\sqrt x}{2\sqrt x}$
$\quad\;\;=\large\frac{\cos\sqrt x}{2\sqrt x}-\frac{\sin 2\sqrt x}{2\sqrt x}$
answered Jun 27, 2013 by sreemathi.v
 
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