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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} $

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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}$
$\Rightarrow y' = \large \frac{1}{cos e^x}$$\times -sin(e^x)\; e^x$
$\Rightarrow y' = - e^x\; \tan{e^x}$
answered Apr 10, 2013 by balaji.thirumalai
 
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