Question

Differentiate ${\sin }^{-1}\sqrt{1-x^2}$ with respect to ${\cos }^{-1}x,\mathrm{if}$
(i) $x\in \left(0,1\right)$
(ii) $x\in \left(-1,0\right)$

Answer 1

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

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=\frac{-1}{\sqrt{1-x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right), \\ v={\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

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


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let},u={\sin }^{-1}\sqrt{1-x^2} \\ \mathrm{Put}x=\cos \theta \\ \Rightarrow u={\sin }^{-1}\sqrt{1-{\cos }^2\theta } \\ \Rightarrow u={\sin }^{-1}\left(\sin \theta \right)...\left(\mathrm{i}\right) \\ \mathrm{And},v={\cos }^{-1}x...\left(\mathrm{ii}\right) \\ \mathrm{Now},x\in \left(-1,0\right) \\ \Rightarrow \cos \theta \in \left(-1,0\right) \\ \Rightarrow \theta \in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\mathrm{\pi }-\theta \left[\mathrm{Since},{\sin }^{-1}\left(\sin \theta \right)=\mathrm{\pi }-\theta \mathrm{if}\theta \in \left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow u=\mathrm{\pi }-{\cos }^{-1}x\left[\mathrm{Since},x=\cos \theta \right] \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0-\frac{-1}{\sqrt{1-x^2}} \\ \Rightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right), \\ v={\cos }^{-1}x \end{gathered}$$

Differentiating it with respect to x,

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