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If Y = Sin [ 2 Tan − 1 { √ 1 − X 1 + X } ] , Find D Y D X ? - Mathematics

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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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पाठ 11: Differentiation - Exercise 11.03 [पृष्ठ ६४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.03 | Q 41 | पृष्ठ ६४

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