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