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Question
\[y = \left( \sin x \right)^{\left( \sin x \right)^{\left( \sin x \right)^{. . . \infty}}} \],prove that \[\frac{y^2 \cot x}{\left( 1 - y \log \sin x \right)}\] ?
Solution
\[{ \text{ We have, y} }= \left( \sin x \right)^{\left( \sin x \right)^{\left( \sin x \right)^{. . . . \infty}}} \]
\[ \Rightarrow y = \left( \sin x \right)^y\]
Taking log on both sides,
\[\log y = \log \left( \sin x \right)^y \]
\[ \Rightarrow \log y = y \log\left( \sin x \right)\]
\[\Rightarrow \frac{1}{y}\frac{dy}{dx} = y\frac{d}{dx}\left\{ \log\left( \sin x \right) \right\} + \log \sin x\frac{dy}{dx}\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = y\left( \frac{1}{\sin x} \right)\frac{d}{dx}\left( \sin x \right) + \log \sin x\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1}{y} - \log \sin x \right) = \frac{y}{\sin x}\left( \cos x \right)\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1 - y \log \sin x}{y} \right) = y \cot x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y^2 \cot x}{\left( 1 - y \log \sin x \right)}\]
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