Question

Differentiate

${\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)$

Answer 1

$\mathrm{Let}y={\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left\{{\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)\right\} \\ =\frac{1}{\sqrt{1-{\left(\frac{x}{\sqrt{x^2+a^2}}\right)}^2}}\times \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+a^2}}\right)\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\mathrm{and}\mathrm{quotient}\mathrm{rule}\right] \\ =\frac{1}{\sqrt{1-{\left(\frac{x}{\sqrt{x^2+a^2}}\right)}^2}}\times \left[\frac{{\left(x^2+a^2\right)}^{{\displaystyle \frac{1}{2}}}{\displaystyle \frac{d}{dx}}\left(x\right)-x{\displaystyle \frac{d}{dx}}{\left(x^2+a^2\right)}^{\frac{1}{2}}}{{\left[{\left(x^2+a^2\right)}^{\frac{1}{2}}\right]}^2}\right] \\ =\frac{\sqrt{x^2+a^2}}{\sqrt{x^2+a^2-x^2}}\left[\frac{\sqrt{x^2+a^2}-{\displaystyle \frac{x}{2\sqrt{x^2+a^2}}}{\displaystyle \frac{d}{dx}}\left(x^2+a^2\right)}{\left(x^2+a^2\right)}\right] \\ =\frac{\sqrt{x^2+a^2}}{a\left(x^2+a^2\right)}\left[\sqrt{x^2+a^2}-\frac{x}{2\sqrt{x^2+a^2}}\times 2x\right] \\ =\frac{\sqrt{x^2+a^2}}{a\left(x^2+a^2\right)}\left[\frac{x^2+a^2-x^2}{\sqrt{x^2+a^2}}\right] \\ =\frac{a^2}{a\left(x^2+a^2\right)} \\ =\frac{a}{\left(x^2+a^2\right)} \\ \mathrm{So},\frac{d}{dx}\left\{{\sin }^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right)\right\}=\frac{a}{\left(x^2+a^2\right)} \end{gathered}$$