Differentiate the cos (log x + ex), x > 0 w.r.t. x. - Vidyadip

Question

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

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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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