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प्रश्न
If \[y = \sin \left[ 2 \tan^{- 1} \left\{ \frac{\sqrt{1 - x}}{1 + x} \right\} \right], \text{ find } \frac{dy}{dx}\] ?
उत्तर
\[\text{Here, y }= \sin\left[ 2 \tan^{- 1} \left( \sqrt{\frac{1 - x}{1 + x}} \right) \right]\]
\[\text{Put x } = \cos 2\theta\]
\[\text{We have, y }= \sin\left[ 2 \tan^{- 1} \left( \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} \right) \right]\]
\[ = \sin\left[ 2 \tan^{- 1} \left( \sqrt{\frac{2 \sin^2 \theta}{2 \cos^2 \theta}} \right) \right]\]
\[ = \sin\left[ 2 \tan^{- 1} \sqrt{\tan^2 \theta} \right] \]
\[ = \sin\left[ 2 \tan^{- 1} \left( \tan\theta \right) \right]\]
\[ = \sin\left( 2\theta \right)\]
\[ = \sin\left[ 2 \times \frac{1}{2} \cos^{- 1} x \right] \left[ \text{ Since, x }= \cos 2\theta \right]\]
\[ = \sin\left( \sin^{- 1} \sqrt{1 - x^2} \right)\]
\[ = \sqrt{1 - x^2}\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = \frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2\sqrt{1 - x^2}}\left( - 2x \right)\]
\[ \therefore \frac{d y}{d x} = \frac{- x}{\sqrt{1 - x^2}}\]
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