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

Answer

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

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