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Differentiate Sin − 1 { √ 1 − X 2 } , 0 < X < 1 ? - Mathematics

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प्रश्न

Differentiate \[\sin^{- 1} \left\{ \sqrt{1 - x^2} \right\}, 0 < x < 1\] ?

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उत्तर

\[\text{ Let, y } = \sin^{- 1} \left\{ \sqrt{1 - x^2} \right\}\]

\[\text{ Put, x } = \cos \theta\]

\[ y = \sin^{- 1} \left\{ \sqrt{1 - \cos^2 \theta} \right\}\]

\[ y = \sin^{- 1} \left( \sin\theta \right) . . . \left( 1 \right)\]

\[\text{ Here }, 0 < x < 1\]

\[ \Rightarrow 0 < \cos \theta < 1\]

\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]

\[\text{ So, from equation} \left( 1 \right), \]

\[ y = \theta \left[ \text{Since }, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if } \theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]

\[ y = \cos^{- 1} x \left[ \text{ Since }, x = \cos \theta \right]\]

\[\text{ Differentiating it with respect to x }, \]

\[\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^2}}\]

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अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.03 | Q 4 | पृष्ठ ६३

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