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