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प्रश्न
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right), \text { if }- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?
उत्तर
\[\text { Let, u } = {\sin^{- 1}} \left( 2x\sqrt{1 - x^2} \right)\]
\[\text { Put x } = \sin\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\sin\theta\sqrt{1 - \sin^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \sin\theta\cos\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)\]
\[\text { Let v } = \tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \frac{\sin\theta}{\sqrt{1 - \sin^2 \theta}} \right) \]
\[ \Rightarrow v = \tan^{- 1} \left( \frac{\sin\theta}{\cos\theta} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \tan\theta \right) . . . \left( ii \right)\]
\[\text { Here }, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\]
\[ \Rightarrow - \frac{1}{\sqrt{2}} < \sin\theta < \frac{1}{\sqrt{2}}\]
\[ \Rightarrow - \frac{\pi}{4} < \theta < \frac{\pi}{4}\]
\[\text { So, from equation } \left( i \right), \]
\[u = 2\theta ........\left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow u = 2 \sin^{- 1} x ........\left[ \text { Since, x } = \sin\theta \right]\]
Differentiating it with respect to x,
\[\frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
From equation (ii),
\[v = \theta ........\left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow v = \sin^{- 1} x ........\left[ \text { Since, x } = \sin\theta \right]\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{\sqrt{1 - x^2}} \times \frac{\sqrt{1 - x^2}}{1}\]
\[ \therefore \frac{du}{dv} = 2\]
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