Question

Differentiate \[\left( \sin^{- 1} x \right)^x\] ?

Answer 1

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