Question

Differentiate \[e^{\sin^{- 1} 2x}\] ?

Answer 1

\[\text{Let} y = e^{\sin^{- 1} 2x} \]

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left( e^{\sin^{- 1} 2x} \right)\]

\[ = e^{\sin^{- 1} 2x} \times \frac{d}{dx}\left( \sin^{- 1} 2x \right) \left[ \text{Using chain rule} \right]\]

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

\[ = \frac{2 e^{\sin^{- 1} 2x}}{\sqrt{1 - 4 x^2}}\]

\[So, \frac{d}{dx} \left( e^{\sin^{- 1} 2x} \right) =\frac{2 e^{\sin^{- 1} 2x}}{\sqrt{1 - 4 x^2}}\]