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Differentiate Sin − 1 √ 1 − X 2 with Respect to Cos − 1 X , If X ∈ ( 0 , 1 ) ? - Mathematics

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प्रश्न

Differentiate  \[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cos^{- 1} x, \text { if}\]\[x \in \left( 0, 1 \right)\]  ?

 

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उत्तर

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

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

\[ \Rightarrow u = \sin^{- 1} \sqrt{1 - \cos^2 \theta}\]

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

\[\text { And,} v = \cos^{- 1} x . . . \left( ii \right)\]

\[\text { Now, x } \in \left( 0, 1 \right)\]

\[ \Rightarrow \cos\theta \in \left( 0, 1 \right)\]

\[ \Rightarrow \theta \in \left( 0, \frac{\pi}{2} \right)\]

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

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

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

Differentiating it with respect to x,

\[\frac{du}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right), \]
\[v = \cos^{- 1} x\]

Differentiating it with respect to x,

\[\frac{dv}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{- 1}{\sqrt{1 - x^2}} \times \frac{\sqrt{1 - x^2}}{- 1}\]
\[ \therefore \frac{du}{dx} = 1\]

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अध्याय 11: Differentiation - Exercise 11.08 [पृष्ठ ११२]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.08 | Q 4.1 | पृष्ठ ११२

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