Differentiate $log (cos e^{x})$ w.r.t to $x$

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Solution

By chain rule , $\frac{d}{d x} (l o g (c o s e^{x}))$
$= \frac{1}{c o s (e^{x})} \times \frac{d}{d x} (c o s e^{x}) = \frac{1}{c o s (e^{x})} . (- s i n e^{x}) . \frac{d}{d n} (e^{x})$
= $\frac{- s i n (e^{x})}{c o s (e^{x})} . e^{x} = - e^{x} t a n (e^{x})$

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