# Differentiate the following w.r.t. $x : \log\;( cos \: e^x )$

$\begin{array}{1 1} - e^x\; \tan{e^x} \\ e^x\; \tan{e^x} \\ e^x\; \tan{e^{-x}} \\ -e^x\; \tan{e^{-x}} \end{array}$

Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(cosx)}{dx} $$=-sinx • \; \large \frac{d(e^{x})}{dx}$$= e^{x}$
• $\; \large \frac{d(\log x)}{dx} $$=\large\frac{1}{x} Given y = \log (\cos e^x) This is of the form y = f(g(x), where g(x) = \cos e^{x}, so we can apply the chain rule of differentiation. Now g(x) = \cos e^x itself is of the form g(x) = f_1(f_2(x)), where f_2 = e^{x}, so let us apply the chain rule and calculate g'(x) According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \; \large \frac{d(e^{x})}{dx}$$= e^{x}$ and $\; \large \frac{d(cosx)}{dx} $$=-sinx \Rightarrow g'(x) = -sin(e^x)\; e^x \; \large \frac{d(\log x)}{dx}$$=\large\frac{1}{x}$