Question

Differentiate

${\left({\sin }^{-1}x\right)}^x$

Answer 1

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