Question

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

Answer 1

\[\text {  Let, u } = \sin^{- 1} \sqrt{1 - x^2}\]
\[\text { Put x} = \cos\theta\]
\[ \Rightarrow u = \sin^{- 1} \sqrt{1 - \cos^2 \theta}\]
\[ \Rightarrow u = \sin^{- 1} \left( \sin\theta \right) . . . \left( i \right)\]
\[\text { And, v} = \cos^{- 1} x . . . \left( ii \right)\]
\[\text { Now, x } \in \left( - 1, 0 \right)\]
\[ \Rightarrow \cos\theta \in \left( - 1, 0 \right)\]
\[ \Rightarrow \theta \in \left( \frac{\pi}{2}, \pi \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = \pi - \theta \left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \pi - \theta if \theta \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \right]\]
\[ \Rightarrow u = \pi - \cos^{- 1} x \left[ \text { Since, x } = \cos\theta \right]\]

Differentiating it with respect to x,

\[\frac{du}{dx} = 0 - \frac{- 1}{\sqrt{1 - x^2}} \]
\[ \Rightarrow \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text { from equation } \left( ii \right), \]
\[v = \cos^{- 1} x\]

Differentiating it with respect to x,

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