Question

Differentiate ${\sin }^{-1}\sqrt{1-x^2}$ with respect to ${\cot }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),\mathrm{if}0<x<1$

Answer 1

$$\begin{gathered} \mathrm{Let},u={\sin }^{-1}\left(\sqrt{1-x^2}\right) \\ \mathrm{Put}x=\cos \theta \\ \Rightarrow \theta ={\cos }^{-1}x \\ \mathrm{We}\mathrm{get},u={\sin }^{-1}\left(\sin \theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Let},v=co{t}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \\ \Rightarrow v=co{t}^{-1}\left(\frac{\cos \theta }{\sqrt{1-{\cos }^2\theta }}\right) \\ \Rightarrow v=co{t}^{-1}\left(\frac{\cos \theta }{\sin \theta }\right) \\ \Rightarrow v=co{t}^{-1}\left(cot\theta \right)...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ 0<x<1 \\ \Rightarrow 0<\cos \theta <1 \\ \Rightarrow 0<\theta <\frac{\mathrm{\pi }}{2} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\theta \left[\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\theta ,\mathrm{if}\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ \Rightarrow u={\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=\frac{-1}{\sqrt{1-x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{From}\mathrm{equation}\left(\mathrm{ii}\right), \\ v=\theta \left[\mathrm{Since},co{t}^{-1}\left(cot\theta \right)=\theta ,\mathrm{if}\theta \in \left(0,\mathrm{\pi }\right)\right] \\ \Rightarrow v={\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

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