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Question
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\sec^{- 1} \left( \frac{1}{\sqrt{1 - x^2}} \right)\], if \[x \in \left( \frac{1}{\sqrt{2}}, 1 \right)\] ?
Solution
\[\text { Let, u }= \sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)\]
\[ \text { Put x } = \sin\theta\]
\[ \Rightarrow u = \sin^{- 1} \left( 2\sin\theta\sqrt{1 - \sin^2 \theta} \right)\]
\[ \Rightarrow u = \sin^{- 1} \left( 2 \sin\theta \cos\theta \right) \]
\[ \Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)\]
\[\text { And, } \]
\[\text { Let, v } = se c^{- 1} \left( \frac{1}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow v = se c^{- 1} \left( \frac{1}{\sqrt{1 - \sin^2 \theta}} \right) \]
\[ \Rightarrow v = se c^{- 1} \left( \frac{1}{\cos\theta} \right) \]
\[ \Rightarrow v = se c^{- 1} \left( sec\theta \right) \]
\[ \Rightarrow v = \cos^{- 1} \left( \frac{1}{\frac{1}{\cos\theta}} \right) \left[ \text { Since }, se c^{- 1} x = \cos^{- 1} \left( \frac{1}{x} \right) \right]\]
\[ \Rightarrow v = \cos^{- 1} \left( \cos\theta \right) . . . \left( ii \right)\]
\[\text { Here }, \]
\[ x \in \left( \frac{1}{\sqrt{2}}, 1 \right)\]
\[ \Rightarrow \sin\theta \in \left( \frac{1}{\sqrt{2}}, 1 \right)\]
\[ \Rightarrow \theta \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)\]
\[\text { So, from equation } \left( i \right), \]
\[ u = 2\theta ........\left[ \text { Since }, \sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right] \]
\[\text { Let, u }= 2 \sin^{- 1} x .........\left[ \text { Since,} x = \sin\theta \right]\]
Differentiating it with respect to x,
\[\frac{du}{dx} = 2\left( \frac{1}{\sqrt{1 - x^2}} \right)\]
\[ \Rightarrow \frac{du}{dx} = \frac{2}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text { And, from equation } \left( ii \right), \]
\[v = \theta \left[ \text{ Since,} \cos^{- 1} \left( \cos\theta \right) = \theta, \text { if } \theta \in \left[ 0, \pi \right] \right]\]
\[ \Rightarrow v = \sin^{- 1} x \left[ \text { Since }, x = \sin\theta \right]\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text {dividing equation } \left( iii \right) by \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{\sqrt{1 - x^2}} \times \frac{\sqrt{1 - x^2}}{1}\]
\[ \therefore \frac{du}{dv} = 2\]
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