Question

Differentiate the following w.r.t. $x$:
$log(cos\,e^x)$

Answer 1

The correct answer is $=-e^xtan\,e^x,e^x≠(2n+1)\frac{π}{2},n∈N$
Let $y=log(cos\,e^x)$
By using the chain rule,we obtain
$\frac{dy}{dx}=\frac{d}{dx}[log(cos\,e^x)]$
$=\frac{1}{cos\,e^x}.\frac{d}{dx}(cos\,e^x)$
$=\frac{1}{cos\,e^x}.(-sin\,e^x).\frac{d}{dx}(e^x)$
$=\frac{-sin\,e^x}{cos\,e^x}.e^x$
$=-e^xtan\,e^x,e^x≠(2n+1)\frac{π}{2},n∈N$