Advertisements
Advertisements
प्रश्न
Differentiate \[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cos^{- 1} x, \text { if}\]\[x \in \left( 0, 1 \right)\] ?
Advertisements
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,
\[\frac{dv}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation } \left( iii \right) by \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{- 1}{\sqrt{1 - x^2}} \times \frac{\sqrt{1 - x^2}}{- 1}\]
\[ \therefore \frac{du}{dx} = 1\]
