Question

Differentiate

$x^{{\cos }^{-1}x}$

Answer 1

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