Question

Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( \frac{1}{2 \sqrt{2}}, \frac{1}{2} \right)\] ?

Answer 1

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

Differentiate it with respect to x,

\[\frac{du}{dx} = 2\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[\frac{du}{dx} = \frac{- 2}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[\frac{du}{dx} = \frac{- 4}{\sqrt{1 - 4 x^2}} . . . \left( iii \right)\]
\[\text { Differentiating equation } \left( ii \right) \text { with respect to x,} \]
\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\frac{d}{dx}\left( 1 - 4 x^2 \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\left( - 8x \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) \text { by } \left( iv \right)\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{- 4}{\sqrt{1 - 4 x^2}} \times \frac{\sqrt{1 - 4 x^2}}{- 4x}\]
\[ \therefore \frac{du}{dv} = \frac{1}{x}\]