# How do you differentiate  y =sin(ln(cos x))  using the chain rule?

Jan 15, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \left(\ln \left(\cos x\right)\right) . \tan x$

#### Explanation:

Differentiate using the 'chain rule' :

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\ln \left(\cos x\right)\right) . \frac{d}{\mathrm{dx}} \left(\ln \left(\cos x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\ln \left(\cos x\right)\right) . \left(\frac{1}{\cos} x . \frac{d}{\mathrm{dx}} \left(\cos x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\ln \left(\cos x\right)\right) . \left(\frac{1}{\cos} x . \left(- \sin x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\ln \left(\cos x\right)\right) . \left(- \sin \frac{x}{\cos} x\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \left(\ln \left(\cos x\right)\right) . \tan x$