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Question
Find \[\frac{dy}{dx}\] \[y = x^{\sin x} + \left( \sin x \right)^x\] ?
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Solution
\[\text{ Let y } = x^{\sin x } + \left( \sin x \right)^x \]
\[\text{ Also, let u } = x^{\sin x } \text{ and v } = \left( \sin x \right)^x \]
\[ \therefore y = u + v\]
\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . \left( i \right)\]
\[\text{ Now, u } = x^{\sin x} \]
\[\text{ Taking log on both sides}, \]
\[ \Rightarrow \log u = \log\left( x^{\sin x} \right)\]
\[ \Rightarrow \log u = \sin x \log x\]
\[\text{ Differentiating both sides with respect to x}, \]
\[\frac{1}{u}\frac{du}{dx} = \log x\frac{d}{dx}\left( \sin x \right) + \sin x\frac{d}{dx}\left( \log x \right) \]
\[ \Rightarrow \frac{du}{dx} = u\left[ \cos x \log x + \sin x\frac{1}{x} \right]\]
\[ \Rightarrow \frac{du}{dx} = x^{\sin x} \left[ \cos x \log x + \frac{\sin x}{x} \right] . . . \left( ii \right)\]
\[\text{ Again, v } = \left( \sin x \right)^x \]
\[\text{ Taking log on both sides }, \]
\[ \Rightarrow \log v = \log \left( \sin x \right)^x \]
\[ \Rightarrow \log v = x \log\left( \sin x \right)\]
\[\text{ Differentiating both sides with respect to x }, \]
\[\frac{1}{v}\frac{dv}{dx} = \log\left( \sin x \right)\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left[ \log\left( \sin x \right) \right]\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ \log\left( \sin x \right) + x\frac{1}{\sin x}\frac{d}{dx}\left( \sin x \right) \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( \sin x \right)^x \left[ \log \sin x + \frac{x}{\sin x}\cos x \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( \sin x \right)^x \left[ \log \sin x + x \cot x \right] . . \left( iii \right)\]
\[\text{ From }\left( i \right), \left( ii \right)\text{ and }\left( iii \right), \text{ we obtain }\]
\[\frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) + \left( \sin x \right)^x \left[ \log \sin x + x \cot x \right] \]
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