Find the derivative of $cos (l o g x + e^{x})$

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Solution

$\frac{d}{d x} cos (log x + e^{x})$
$= - sin (log x + e^{x}) (\frac{1}{x} + e^{x})$
$∴ \frac{d}{d x} cos (log x + e^{x}) = - \frac{sin (log x + e^{x}) (1 + x e^{x})}{x}$

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