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Question
Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2}, - \frac{1}{2 \sqrt{2}} \right)\] ?
Solution
\[\text { Let, u } = \sin^{- 1} \left( 4x\sqrt{1 - 4 x^2} \right)\]
\[ \text { put,} 2x = \cos\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \times \cos\theta\sqrt{1 - \cos^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\cos\theta \sin\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin 2\theta \right) . . . \left( i \right)\]
\[ \text {Let, v } = \sqrt{1 - 4 x^2} . . . \left( ii \right)\]
\[\text { Here }, \]
\[ x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[ \Rightarrow 2x \in \left( - 1, - \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow \theta \in \left( \frac{3\pi}{4}, \pi \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = \pi - 2\theta ......\left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \pi - \theta , \text{ if }\theta \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \right]\]
\[ \Rightarrow u = \pi - 2 \cos^{- 1} \left( 2x \right) ........\left[ \text { Since }, 2x = \cos\theta \right]\]
Differentiate it with respect to x,
\[\frac{du}{dx} = 0 - 2\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{4}{\sqrt{1 - 4 x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right), \]
\[\frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}}\]
\[\text { but }, x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[ \therefore \frac{dv}{dx} = \frac{- 4\left( - x \right)}{\sqrt{1 - 4 \left( - x \right)^2}}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{4x}{\sqrt{1 - 4 x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( iv \right)\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{4}{\sqrt{1 - 4 x^2}} \times \frac{\sqrt{1 - 4 x^2}}{4x}\]
\[ \therefore \frac{du}{dv} = \frac{1}{x}\]
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