Question

Differentiate ${\sin }^{-1}\left(4x\sqrt{1-4x^2}\right)$ with respect to $\sqrt{1-4x^2}$, if

(i) $x\in \left(-\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2\sqrt{2}}}\right)$

(ii) $x\in \left(\frac{1}{2\sqrt{2}},\frac{1}{2}\right)$

(iii) $x\in \left(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\right)$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let},u={\sin }^{-1}\left(4x\sqrt{1-4x^2}\right) \\ \mathrm{put}2x=\cos \theta \\ \Rightarrow u={\sin }^{-1}\left(2\times \cos \theta \sqrt{1-{\cos }^2\theta }\right) \\ \Rightarrow u={\sin }^{-1}\left(2\cos \theta \sin \theta \right) \\ \Rightarrow u={\sin }^{-1}\left(\sin 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Let},v=\sqrt{1-4x^2}...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ x\in \left(-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}\right) \\ \Rightarrow 2x\in \left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \\ \Rightarrow \theta \in \left(\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4}\right) \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\mathrm{\pi }-2\mathrm{\theta }\left[\mathrm{Since},{\sin }^{-1}\left(\mathrm{sin\theta }\right)=\mathrm{\pi }-\mathrm{\theta },\mathrm{if}\mathrm{\theta }\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\right] \\ \Rightarrow \mathrm{u}=\mathrm{\pi }-2{\cos }^{-1}\left(2\mathrm{x}\right)\left[\mathrm{Since},2x=\cos \theta \right] \end{gathered}$$

Differentiating it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0-2\left(\frac{-1}{\sqrt{1-{\left(2x\right)}^2}}\right)\frac{d}{dx}\left(2x\right) \\ \Rightarrow \frac{du}{dx}=\frac{2}{\sqrt{1-4x^2}}\left(2\right) \\ \Rightarrow \frac{du}{dx}=\frac{4}{\sqrt{1-4x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right) \\ \frac{dv}{dx}=\frac{-4x}{\sqrt{1-4x^2}} \\ \mathrm{but},x\in \left(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\right) \\ \frac{dv}{dx}=\frac{-4\left(-x\right)}{\sqrt{1-4{\left(-x\right)}^2}} \\ \Rightarrow \frac{dv}{dx}=\frac{4x}{\sqrt{1-4x^2}}...\left(\mathrm{iv}\right) \\ \mathrm{Diferentiating}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}\frac{d}{dx}\left(1-4x^2\right) \\ \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}\left(-8x\right) \\ \Rightarrow \frac{dv}{dx}=\frac{-4x}{\sqrt{1-4x^2}}...\left(\mathrm{v}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{v}\right) \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{4}{\sqrt{1-4x^2}}\times \frac{\sqrt{1-4x^2}}{-4x} \\ \therefore \frac{du}{dv}=-\frac{1}{x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let},u={\sin }^{-1}\left(4x\sqrt{1-4x^2}\right) \\ \mathrm{put}2x=\cos \theta \\ u={\sin }^{-1}\left(2\times \cos \theta \sqrt{1-{\cos }^2\theta }\right) \\ \Rightarrow u={\sin }^{-1}\left(2\cos \theta \sin \theta \right) \\ \Rightarrow u={\sin }^{-1}\left(\sin 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Let},v=\sqrt{1-4x^2}...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ x\in \left(\frac{1}{2\sqrt{2}},\frac{1}{2}\right) \\ \Rightarrow 2x\in \left(\frac{1}{\sqrt{2}},1\right) \\ \Rightarrow \cos \theta \in \left(\frac{1}{\sqrt{2}},1\right) \\ \Rightarrow \theta \in \left(0,\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=2\mathrm{\theta }\left[\mathrm{Since},{\sin }^{-1}\left(\mathrm{sin\theta }\right)=\mathrm{\theta },\mathrm{if}\mathrm{\theta }\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow \mathrm{u}=2{\cos }^{-1}\left(2\mathrm{x}\right)\left[\mathrm{Since},2x=\cos \theta \right] \end{gathered}$$

Differentiate it with respect to x,

$$\begin{gathered} \frac{du}{dx}=2\left(\frac{-1}{\sqrt{1-{\left(2x\right)}^2}}\right)\frac{d}{dx}\left(2x\right) \\ \frac{du}{dx}=\frac{-2}{\sqrt{1-4x^2}}\left(2\right) \\ \frac{du}{dx}=\frac{-4}{\sqrt{1-4x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{Diferentiating}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}\frac{d}{dx}\left(1-4x^2\right) \\ \Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{1-4x^2}}\left(-8x\right) \\ \Rightarrow \frac{dv}{dx}=\frac{-4x}{\sqrt{1-4x^2}}...\left(\mathrm{iv}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{iv}\right) \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{-4}{\sqrt{1-4x^2}}\times \frac{\sqrt{1-4x^2}}{-4x} \\ \therefore \frac{du}{dv}=\frac{1}{x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let},u={\sin }^{-1}\left(4x\sqrt{1-4x^2}\right) \\ \mathrm{put},2x=\cos \theta \\ \Rightarrow u={\sin }^{-1}\left(2\times \cos \theta \sqrt{1-{\cos }^2\theta }\right) \\ \Rightarrow u={\sin }^{-1}\left(2\cos \theta \sin \theta \right) \\ \Rightarrow u={\sin }^{-1}\left(\sin 2\theta \right)...\left(\mathrm{i}\right) \\ \mathrm{Let},v=\sqrt{1-4x^2}...\left(\mathrm{ii}\right) \\ \mathrm{Here}, \\ x\in \left(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\right) \\ \Rightarrow 2x\in \left(-1,-\frac{1}{\sqrt{2}}\right) \\ \Rightarrow \theta \in \left(\frac{3\mathrm{\pi }}{4},\mathrm{\pi }\right) \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ u=\mathrm{\pi }-2\mathrm{\theta }\left[\mathrm{Since},{\sin }^{-1}\left(\mathrm{sin\theta }\right)=\mathrm{\pi }-\mathrm{\theta },\mathrm{if}\mathrm{\theta }\in \left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\right)\right] \\ \Rightarrow \mathrm{u}=\mathrm{\pi }-2{\cos }^{-1}\left(2\mathrm{x}\right)\left[\mathrm{Since},2x=\cos \theta \right] \end{gathered}$$


Differentiate it with respect to x,

$$\begin{gathered} \frac{du}{dx}=0-2\left(\frac{-1}{\sqrt{1-{\left(2x\right)}^2}}\right)\frac{d}{dx}\left(2x\right) \\ \Rightarrow \frac{du}{dx}=\frac{2}{\sqrt{1-4x^2}}\left(2\right) \\ \Rightarrow \frac{du}{dx}=\frac{4}{\sqrt{1-4x^2}}...\left(\mathrm{iii}\right) \\ \mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right), \\ \frac{dv}{dx}=\frac{-4x}{\sqrt{1-4x^2}} \\ \mathrm{but},x\in \left(-\frac{1}{2},-\frac{1}{2\sqrt{2}}\right) \\ \therefore \frac{dv}{dx}=\frac{-4\left(-x\right)}{\sqrt{1-4{\left(-x\right)}^2}} \\ \Rightarrow \frac{dv}{dx}=\frac{4x}{\sqrt{1-4x^2}}...\left(\mathrm{iv}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{by}\left(\mathrm{iv}\right) \\ \frac{{\displaystyle \frac{du}{dx}}}{{\displaystyle \frac{dv}{dx}}}=\frac{4}{\sqrt{1-4x^2}}\times \frac{\sqrt{1-4x^2}}{4x} \\ \therefore \frac{du}{dv}=\frac{1}{x} \end{gathered}$$