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प्रश्न
Differentiate \[\tan^{- 1} \left\{ \frac{x}{a + \sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?
उत्तर
\[\text{ Let, y } = \tan^{- 1} \left\{ \frac{x}{a + \sqrt{a^2 - x^2}} \right\}\]
\[\text{ Put x }= a \sin\theta\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a + \sqrt{a^2 - a^2 \sin^2 \theta}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{a \sin\theta}{a + \sqrt{a^2 \left( 1 - \sin^2 \theta \right)}} \right) \]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a + a \cos\theta} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{a \sin\theta}{a\left( 1 + \cos\theta \right)} \right\} \]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{\sin\theta}{1 + \cos\theta} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right\}\]
\[ \Rightarrow y = \tan^{- 1} \left( \tan \frac{\theta}{2} \right) . . . \left( i \right) \]
\[\text{Here }, - a < x < a\]
\[ \Rightarrow - 1 < \frac{x}{a} < 1\]
\[ \Rightarrow - 1 < \sin\theta < 1\]
\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2}\]
\[ \Rightarrow - \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\]
\[\text{ So, from equation } \left( i \right), \]
\[ y = \frac{\theta}{2} .......\left[ \text{ Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ \Rightarrow y = \frac{1}{2} \sin^{- 1} \left( \frac{x}{a} \right) ..........\left[ \text{ Since }, x = a \sin\theta \right]\]
\[\text{ Differentiating it with respect to x }, \]
\[ \frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}}\frac{d}{dx}\left( \frac{x}{a} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{a}{2\sqrt{a^2 - x^2}} \times \left( \frac{1}{a} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{1}{2\sqrt{a^2 - x^2}}\]
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