Differentiate the following functions with respect to x :

Solution:

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

Taking log both the sides:

$\Rightarrow \log y=\log \left(\sin ^{-1} x\right)^{x}$

$\Rightarrow \log y=x \log \left(\sin ^{-1} x\right)\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x} \log \left(\sin ^{-1} \mathrm{x}\right)\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}\left(\log \left(\sin ^{-1} \mathrm{x}\right)\right)}{\mathrm{dx}}+\log \left(\sin ^{-1} \mathrm{x}\right) \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=x \times \frac{1}{\sin ^{-1} x} \frac{d\left(\sin ^{-1} x\right)}{d x}+\log \left(\sin ^{-1} x\right)$

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{x}{\sin ^{-1} x} \times \frac{1}{\sqrt{1-x^{2}}} \frac{d x}{d x}+\log \left(\sin ^{-1} x\right)$

$\left\{\frac{d\left(\sin ^{-1} u\right)}{d x}=\frac{1}{\sqrt{1-u^{2}}} \frac{d u}{d x}\right\}$

$\left\{\frac{\mathrm{d}\left(\sin ^{-1} \mathrm{u}\right)}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{x}{\sin ^{-1} x \sqrt{1-x^{2}}}+\log \left(\sin ^{-1} x\right)$

$\Rightarrow \frac{d y}{d x}=y\left\{\frac{x}{\sin ^{-1} x \sqrt{1-x^{2}}}+\log \left(\sin ^{-1} x\right)\right\}$

Put the value of $y=\left(\sin ^{-1} x\right)^{x}$ :

$\left.\Rightarrow \frac{d y}{d x}=\left(\sin ^{-1} x\right)^{x}\left\{\frac{x}{\sin ^{-1} x \sqrt{1-x^{2}}}+\log \left(\sin ^{-1} x\right)\right\}\right\}$