Question

Differentiate: $\log \left(\cos e^x\right)$

Answer 1

We will apply chain rule.

Let $\cos u=t$

$u=e^x$

$\frac{d}{dx}\log t=\frac{1}{t}\frac{dt}{dx}=\frac{1}{t}\frac{d\left(\cos u\right)}{dx}=-\frac{1}{t}\sin u\frac{du}{dx}=-\frac{1}{t}\sin u\frac{d{e}^x}{dx}=-\frac{1}{t}\sin u{e}^x$

Substituting all the values will give

$\frac{d}{dx}\log t=-e^x\frac{1}{\cos e^x}\sin e^x=-e^x\tan e^x$