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प्रश्न
Differentiate \[\tan^{- 1} \left\{ \frac{x}{\sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?
उत्तर
\[\text{Let, y } = \tan^{- 1} \left\{ \frac{x}{\sqrt{a^2 - x^2}} \right\}\]
\[\text{ Put x } = a \sin\theta\]
\[ y = \tan^{- 1} \left\{ \frac{a \sin\theta}{\sqrt{a^2 - a^2 \sin^2 \theta}} \right\}\]
\[ y = \tan^{- 1} \left( \frac{a \sin\theta}{\sqrt{a^2 \left( 1 - \sin^2 \theta \right)}} \right) \]
\[ y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a \cos\theta} \right\} \]
\[ y = \tan^{- 1} \left( \tan\theta \right) ...........\left( 1 \right)\]
\[Here, - a < x < a\]
\[ \Rightarrow - 1 < \frac{x}{a} < 1\]
\[ \Rightarrow \sin\left( - \frac{\pi}{2} \right) < \sin\theta < \sin\left( \frac{\pi}{2} \right) \left( \because x = a \sin\theta \right)\]
\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2}\]
\[\text{ So, from equation } \left( 1 \right), \]
\[ y = \theta \left[ \text{ Since}, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if } \theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ y = \sin^{- 1} \left( \frac{x}{a} \right) \left[ \text{ Since, x } = a \sin\theta \right]\]
\[\text{ Differentiating it with respect to x }, \]
\[\text{ Using chain rule }, \]
\[\frac{d y}{d x} = \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}}\frac{d}{dx}\left( \frac{x}{a} \right)\]
\[\frac{d y}{d x} = \frac{a}{\sqrt{a^2 - x^2}} \times \left( \frac{1}{a} \right)\]
\[\frac{d y}{d x} = \frac{1}{\sqrt{a^2 - x^2}}\]
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