Differentiate the given function w.r.t. $x$ .
$cos (log x + e^{x}), x > 0$

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Solution

Let $y = cos (log x + e^{x}), x > 0$
Thus using chain rule,
$\frac{d y}{d x} = - sin (log x + e^{x}) . \frac{d}{d x} (log x + e^{x})$
$= - sin (log x + e^{x}) . [\frac{d}{d x} (log x) + \frac{d}{d x} (e^{x})]$
$= - sin (log x + e^{x}) . (\frac{1}{x} + e^{x})$
$= - (\frac{1}{x} + e^{x}) sin (log x + e^{x}), x > 0$

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