Question

Differentiate the following w.r.t. $x$:
$cos(log\,x+e^x),x>0$

Answer 1

The correct answer is $=-(\frac{1}{x}+e^x)sin(log\,x+e^x),x>0$
Let $y=cos(log\,x+e^x)$
By using the chain rule,we obtain
$\frac{dy}{dx}=-sin(log\,x+e^x).\frac{d}{dx}(log\,x+e^x)$
$-sin(log\,x+e^x)[.\frac{d}{dx}(log\,x)+\frac{d}{dx}(e^x)]$
$=-sin(log\,x+e^x).(\frac{1}{x}+e^x)$
$=-(\frac{1}{x}+e^x)sin(log\,x+e^x),x>0$