Question

find dy/dx: y=cos√sinx

Answer 1

$y = \cos( \sqrt{ \sin(x) } ) \\ \frac{dy}{dx} = - (\sin( \sqrt[2]{sinx} ) . \frac{1}{2 \sqrt{ \sin(x) } } . \cos(x)$
hope it will help u...

Answer 2

Answer:

$-\sin (\sqrt{\sin x})\frac{1}{2\sqrt{\sin x}}cos x$

Step-by-step explanation:

Given equation,

$y=\cos\sqrt{ \sin x }$

Differentiating with respect to x,

$\frac{dy}{dx}=-\sin (\sqrt{\sin x})\frac{d}{dx}(\sqrt{\sin x})$

$(\because \frac{d}{dx} \cos x =-\sin x)$

$\frac{dy}{dx}=-\sin (\sqrt{\sin x})\frac{1}{2\sqrt{\sin x}}\frac{d}{dx}(\sin x)$

$(\because \frac{d}{dx} \sqrt{x}=\frac{1}{2\sqrt{x}})$

$\frac{dy}{dx}=-\sin (\sqrt{\sin x})\frac{1}{2\sqrt{\sin x}}cos x$

$(\because \frac{d}{dx} \sin x =\cos x)$