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Question
Differentiate \[\sin \left( x^x \right)\] ?
Solution
\[\text{ Let y }= \sin x^x \]
\[ \Rightarrow \sin^{- 1} y = x^x . . . \left( i \right)\]
\[\text{ Taking log on both sides}, \]
\[\log\left( \sin^{- 1} y \right) = \log x^x \]
\[ \Rightarrow \log\left( \sin^{- 1} y \right) = x \log x \]
\[\text{ Differentiating with respect to x }, \]
\[ \Rightarrow \frac{1}{\sin^{- 1} y}\frac{dy}{dx}\left( \sin^{- 1} y \right) = x\frac{d}{dx}\log x + \log x\frac{d}{dx}x \]
\[ \Rightarrow \frac{1}{\sin^{- 1} y} \times \left( \frac{1}{\sqrt{1 - y^2}} \right)\frac{dy}{dx} = x\left( \frac{1}{x} \right) + \log x\]
\[ \Rightarrow \frac{dy}{dx} = \sin^{- 1} y\sqrt{1 - y^2}\left( 1 + \log x \right)\]
\[ \Rightarrow \frac{dy}{dx} = \sin^{- 1} \left( \sin x^x \right)\sqrt{1 - \left( \sin x^x \right)^2}\left( 1 + \log x \right)\]
\[ \therefore \frac{dy}{dx} = x^x \cos x^x \left( 1 + \log x \right) \left[ \text{ using equation } \left( i \right) \right]\]
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