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प्रश्न
उत्तर
\[\text{ Let, u } = \sin^{- 1} \left( 4x\sqrt{1 - 4 x^2} \right)\]
\[ \text { put } 2x = \cos\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \times \cos\theta\sqrt{1 - \cos^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\cos\theta \sin\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin 2\theta \right) . . . \left( i \right)\]
\[ \text { Let, v }= \sqrt{1 - 4 x^2} . . . \left( ii \right)\]
\[\text{Here}, \]
\[ x \in \left( - \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}} \right)\]
\[ \Rightarrow 2x \in \left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow \theta \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = \pi - 2\theta .....\left[ \text { Since }, \sin^{- 1} \left( sin \theta \right) = \pi - \theta , \text{ if }\theta \in \left( \frac{\pi}{2}, \pi \right) \right]\]
\[ \Rightarrow u = \pi - 2 \cos^{- 1} \left( 2x \right) \left[ \text { Since}, 2x = \cos\theta \right]\]
Differentiating it with respect to x,
\[\frac{du}{dx} = 0 - 2\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{4}{\sqrt{1 - 4 x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right)\]
\[\frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}}\]
\[\text { but,} x \in \left( - \frac{1}{2}, - \frac{1}{2\sqrt{2}} \right)\]
\[\frac{dv}{dx} = \frac{- 4\left( - x \right)}{\sqrt{1 - 4 \left( - x \right)^2}}\]
\[ \Rightarrow \frac{dv}{dx} = \frac{4x}{\sqrt{1 - 4 x^2}} . . . \left( iv \right)\]
\[\text { Diferentiating equation } \left( ii \right) \text { with respect to x }, \]
\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\frac{d}{dx}\left( 1 - 4 x^2 \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{1}{2\sqrt{1 - 4 x^2}}\left( - 8x \right)\]
\[ \Rightarrow \frac{dv}{dx} = \frac{- 4x}{\sqrt{1 - 4 x^2}} . . . \left( v \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( v \right)\]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{4}{\sqrt{1 - 4 x^2}} \times \frac{\sqrt{1 - 4 x^2}}{- 4x}\]
\[ \therefore \frac{du}{dv} = - \frac{1}{x}\]
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