Question

Differentiate ${\sin }^{-1}\left(2ax\sqrt{1-a^2{x}^2}\right)$ with respect to $\sqrt{1-a^2{x}^2},\mathrm{if}-\frac{1}{\sqrt{2}}<ax<\frac{1}{\sqrt{2}}$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\sin }^{-1}\left(2ax\sqrt{1-a^2{x}^2}\right) \\ \mathrm{Put}ax=\sin \theta \Rightarrow \theta ={\sin }^{-1}\left(ax\right) \\ \therefore u={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right) \\ \Rightarrow u={\sin }^{-1}\left(2\sin \theta \cos \theta \right) \\ \Rightarrow u={\sin }^{-1}\left(\sin 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{And} \\ \mathrm{Let}, \\ v=\sqrt{1-a^2{x}^2} \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dv}{dx}=\frac{1}{2\sqrt{1-a^2{x}^2}}\times \frac{d}{dx}\left(1-a^2{x}^2\right) \\ \Rightarrow \frac{dv}{dx}=\left(\frac{0-2a^{2}x}{2\sqrt{1-a^2{x}^2}}\right) \\ \Rightarrow \frac{dv}{dx}=\frac{-a^{2}x}{\sqrt{1-a^2{x}^2}}...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ -\frac{1}{\sqrt{2}}<ax<\frac{1}{\sqrt{2}} \\ \Rightarrow -\frac{1}{\sqrt{2}}<\sin \theta <\frac{1}{\sqrt{2}} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\theta <\frac{\mathrm{\pi }}{4} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=2\theta \left[\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\theta ,\mathrm{if}\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ \Rightarrow u=2{\sin }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=2\times \frac{1}{\sqrt{1-{\left(ax\right)}^2}}\frac{d}{dx}\left(ax\right) \\ \Rightarrow \frac{du}{dx}=\frac{2}{1-a^2{x}^2}\left(a\right) \\ \Rightarrow \frac{du}{dx}=\frac{2a}{1-a^2{x}^2}...\left(\mathrm{iii}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{ii}\right), \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\left(\frac{2a}{\sqrt{1-a^2{x}^2}}\right)\left(\frac{\sqrt{1-a^2{x}^2}}{-a^{2}x}\right) \\ \therefore \frac{du}{dv}=-\frac{2}{ax} \end{gathered}$$