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Question
If \[\left( \cos x \right)^y = \left( \tan y \right)^x\] , prove that \[\frac{dy}{dx} = \frac{\log \tan y + y \tan x}{ \log \cos x - x \sec y \ cosec\ y }\] ?
Solution
\[\text{ We have,} \left( \cos x \right)^y = \left( \tan y \right)^x\]
Taking log on both sides,]
\[\log \left( \cos x \right)^y = \log \left( \tan y \right)^x \]
\[ \Rightarrow y \log \cos x = x \log \tan y\]
Differentiating it with respect to x using chain,
\[\frac{d}{dx}\left( y \log \cos x \right) = \frac{d}{dx}\left( x \log \tan y \right)\]
\[ \Rightarrow y\frac{d}{dx}\left( \log \cos x \right) + \log \cos x\frac{dy}{dx} = x\frac{d}{dx}\left( \log \tan y \right) + \log \tan y\frac{d}{dx}\left( x \right)\]
\[ \Rightarrow y\frac{1}{\cos x}\frac{d}{dx}\left( \cos x \right) + \log \cos x\frac{dy}{dx} = x\frac{1}{\tan y}\frac{d}{dx}\left( \tan y \right) + \log \tan y\]
\[ \Rightarrow \frac{y}{\cos x}\left( - \sin x \right) + \log \cos x\frac{dy}{dx} = \left\{ \frac{x}{\tan y}\left( \sec^2 y \right) \right\}\frac{dy}{dx} + \log \tan y\]
\[ \Rightarrow - y\tan x + \log \cos x\frac{dy}{dx} = \sec y \ cosec\ y \times x\frac{dy}{dx} + \log \tan y\]
\[ \Rightarrow \frac{dy}{dx}\left[ \log \cos x - x \sec y \ cose c \ y \right ] = \log \tan y + y \tan x\]
\[ \Rightarrow \frac{dy}{dx} = \left[ \frac{\log \tan y + y \tan x}{\log \cos x - x\sec y\ cosec\ y } \right]\]
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