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