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Find $\large\frac{dy}{dx}$ of the function $(\cos x)^{\large y} = (\cos y)^{\large x}$

$\begin{array}{1 1}\large\frac{y\tan x+\log\cos y}{x\tan y+\log \cos x} \\ \large\frac{y\tan x-\log\cos y}{x\tan y+\log \cos x} \\ \large\frac{x\tan x+\log\cos y}{x\tan y+\log \cos x} \\ \large\frac{y\tan x+\log\cos y}{x\tan y+\log \sin x} \end{array} $

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  • $(uv)'=u'v+uv'$
  • $\log m^{\large n}=n\log m$
Step 1:
$ (\cos x)^{\large y} = (\cos y)^{\large x} $
Taking $\log$ on both sides
$ \log(\cos x)^{\large y} =\log (\cos y)^{\large x} $
Differentiating with respect to $x$
$\log m^{\large n}=n\log m$
$y\log(\cos x)=x\log(\cos y)$
$\large\frac{dy}{dx}$$\log(\cos x)+y.\large\frac{1}{\cos x}\frac{d}{dx}$$(\cos x)=1.\log\cos y+x.\large\frac{1}{\cos y}\frac{d}{dx}$$\cos y$
Step 2:
$\log(\cos x)\large\frac{dy}{dx}+\frac{y}{\cos x}$$(-\sin x)=\log(\cos y)+\big(\large\frac{-\sin y}{\cos y}\big)\frac{dy}{dx}$
$[\log(\cos x)+x\tan y]\large\frac{dy}{dx}$$=\log\cos y+y\tan x$
$\large\frac{dy}{dx}=\large\frac{y\tan x+\log\cos y}{x\tan y+\log \cos x}$
answered May 9, 2013 by sreemathi.v
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