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Question
If \[y = \sin \left( x^x \right)\] prove that \[\frac{dy}{dx} = \cos \left( x^x \right) \cdot x^x \left( 1 + \log x \right)\] ?
Solution
\[\text{ Let y} = \sin\left( x^x \right) . . . \left( i \right)\]
\[\text{ Also, Let u} = x^x . . . \left( ii \right)\]
\[\text{ Taking log on both sides}, \]
\[ \Rightarrow \log u = \log x^x \]
\[ \Rightarrow \log u = x\log x\]
Differentiating both sides with respect to x,
\[\frac{1}{u}\frac{du}{dx} = \frac{d}{dx}\left( x \log x \right)\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}x\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = x\left( \frac{1}{x} \right) + \log x\left( 1 \right)\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = 1 + \log x\]
\[ \Rightarrow \frac{du}{dx} = u\left( 1 + \log x \right)\]
\[ \Rightarrow \frac{du}{dx} = x^x \left( 1 + \log x \right) . . . \left( iii \right) \left[ \text{ using equation }\left( ii \right) \right]\]
\[\text{ Now, using equation} \left( ii \right) \text{ in equation} \left( i \right), \]
\[y = \sin u\]
\[\text{ Differentiating with respect to x,} \]
\[\frac{dy}{dx} = \frac{d}{dx}\left( \sin u \right)\]
\[ \Rightarrow \frac{dy}{dx} = \cos u\frac{du}{dx}\]
\[\text{ Using equation} \left( ii \right)\text{ and } \left( iii \right), \]
\[\frac{dy}{dx} = \cos\left( x^x \right) \times x^x \left( 1 + \log x \right)\]
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