Question

Differentiate \[\sin^{- 1} \left( 2 ax \sqrt{1 - a^2 x^2} \right)\] with respect to \[\sqrt{1 - a^2 x^2}, \text{ if }-\frac{1}{\sqrt{2}} < ax < \frac{1}{\sqrt{2}}\] ?

Answer 1

\[\text { Let, u }= \sin^{- 1} \left( 2ax\sqrt{1 - a^2 x^2} \right)\]

\[\text { Put ax } = \sin\theta \Rightarrow \theta = \sin^{- 1} \left( ax \right)\]

\[ \therefore u = \sin^{- 1} \left( 2\sin\theta\sqrt{1 - \sin^2 \theta} \right)\]

\[ \Rightarrow u = \sin^{- 1} \left( 2 \sin\theta\cos\theta \right)\]

\[ \Rightarrow u = \sin^{- 1} \left( \sin2\theta \right) . . . \left( i \right)\]

\[\text { And }\]

\[\text { Let,} \]

\[ v = \sqrt{1 - a^2 x^2}\]

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

\[\frac{dv}{dx} = \frac{1}{2\sqrt{1 - a^2 x^2}} \times \frac{d}{dx}\left( 1 - a^2 x^2 \right) \]

\[ \Rightarrow \frac{dv}{dx} = \left( \frac{0 - 2 a^2 x}{2\sqrt{1 - a^2 x^2}} \right) \]

\[ \Rightarrow \frac{dv}{dx} = \frac{- a^2 x}{\sqrt{1 - a^2 x^2}} . . . \left( ii \right)\]

\[\text { Here }, \]

\[ - \frac{1}{\sqrt{2}} < ax < \frac{1}{\sqrt{2}}\]

\[ \Rightarrow - \frac{1}{\sqrt{2}} < \sin\theta < \frac{1}{\sqrt{2}}\]

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

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

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

\[ \Rightarrow u = 2 \sin^{- 1} x\]

Differentiating it with respect to x,

\[\frac{du}{dx} = 2 \times \frac{1}{\sqrt{1 - \left( ax \right)^2}}\frac{d}{dx}\left( ax \right) \]

\[ \Rightarrow \frac{du}{dx} = \frac{2}{1 - a^2 x^2}\left( a \right) \]

\[ \Rightarrow \frac{du}{dx} = \frac{2a}{1 - a^2 x^2} . . . \left( iii \right) \]

\[\text { Dividing equation } \left( iii \right) by \left( ii \right), \]

\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \left( \frac{2a}{\sqrt{1 - a^2 x^2}} \right)\left( \frac{\sqrt{1 - a^2 x^2}}{- a^2 x} \right)\]

\[ \therefore \frac{du}{dv} = - \frac{2}{ax}\]