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

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Question

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

Sum
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Solution

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

\[\text{ Put x  }= \cos 2\theta\]

\[ y = \sin^{- 1} \left\{ \sqrt{\frac{1 - \cos2\theta}{2}} \right\}\]

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

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

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

\[ \Rightarrow 0 < \cos 2\theta < 1\]

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

\[ \Rightarrow 0 < \theta < \frac{\pi}{4}\]

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

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

\[ y = \frac{1}{2} \cos^{- 1} x \left[ \text{ Since, x } = \cos 2\theta \right]\]

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

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

\[\]

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Chapter 11: Differentiation - Exercise 11.03 [Page 63]

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RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.03 | Q 3 | Page 63

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