$\cos (\log x + {e^x}),x > 0$

Let$y = \cos (\log x + {e^x})$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\left\{ {\cos (\log x + {e^x})} \right\}$

$= - \sin (\log x + {e^x})\cfrac{d}{{dx}}(\log x + {e^x})$
$= - \sin (\log x + {e^x})\left[ {\cfrac{1}{x} + {e^x}} \right]$

$= - \left( {\cfrac{1}{x} + {e^x}} \right)\sin (\log x + {e^x}),x > 0$

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