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If $f(x)=e^{\sin (\log \cos x)}$ and $g(x)=\log \cos x$, then what is the derivative of $\mathrm{f}(\mathrm{x})$ with respect to $\mathrm{g}(\mathrm{x})$?
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Verified Answer
The correct answer is:
$\mathrm{f}(\mathrm{x}) \cos [\mathrm{g}(\mathrm{x})]$
Given funcion is: $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\sin (\log \cos \mathrm{x})}$
Differentiating w.r.t. $\mathrm{x}$
$f^{\prime}(x)=e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \frac{1}{\cos x}(-\sin x)$
$=-\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x$
and $\mathrm{g}(\mathrm{x})=\log \cos \mathrm{x}$
$\therefore g^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$
Hence,
$\frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x}{-\tan x}$
$=\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x)$
$=\mathrm{f}(\mathrm{x}) \cdot \cos [\mathrm{g}(\mathrm{x})]$
Differentiating w.r.t. $\mathrm{x}$
$f^{\prime}(x)=e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \frac{1}{\cos x}(-\sin x)$
$=-\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x$
and $\mathrm{g}(\mathrm{x})=\log \cos \mathrm{x}$
$\therefore g^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$
Hence,
$\frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x}{-\tan x}$
$=\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x)$
$=\mathrm{f}(\mathrm{x}) \cdot \cos [\mathrm{g}(\mathrm{x})]$
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