Question
Differentiate the $\cos \left(\log x+e^x\right), x>0$ w.r.t. $x$.

Answer

Let y $= \cos(logx + e^x)$
$\therefore \frac{{dy}}{{dx}} = - \sin \left( {\log x + {e^x}} \right)\frac{d}{{dx}}\left( {\log x + {e^x}} \right)$
$ = - \sin \left( {\log x + {e^x}} \right).\left( {\frac{1}{x} + {e^x}} \right)$

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