Hi,

I'm not arriving at the correct answer, kindly point out my mistake(s).

Derivative of:

$\displaystyle \displaystyle y = \sin (\cos\,x)$

$\displaystyle \displaystyle u = \cos\,x$

$\displaystyle \displaystyle \frac{du}{dx} = -\sin\,x$

$\displaystyle \displaystyle v = \sin\,u$

$\displaystyle \displaystyle \frac{dv}{du} = \cos\,u $

$\displaystyle \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}$

$\displaystyle \displaystyle = \sin\,u(-\sin\,x) + \cos\,x(\cos\,u)$

$\displaystyle \displaystyle = -\sin^{2}x[\cos(x)] + \cos x [\cos(\cos x)]$