# How do you differentiate f(x)=ln(cos(e^(x) ))  using the chain rule?

Mar 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan {e}^{x} \cdot {e}^{x}$

#### Explanation:

So, we got three functions here:

$\ln \left(\cos \left({e}^{x}\right)\right)$

$\cos \left({e}^{x}\right)$

and

${e}^{x}$

Let $y = \ln \left(\cos \left({e}^{x}\right)\right)$

differentiating w.r.t. $x$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \cos \left({e}^{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} \left({e}^{x}\right) \cdot \sin \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} \left({e}^{x}\right) \cdot \sin \left({e}^{x}\right) \cdot {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \frac{{e}^{x}}{\cos} \left({e}^{x}\right) \cdot {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan {e}^{x} \cdot {e}^{x}$

This will be the differentiated function.