If \(y = \sin^{-1}(2x)\), \(-\displaystyle\frac{1}{2} \lt x \lt \frac{1}{2}\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{\sqrt{1 - 4x^{2}}}\)
Since \(\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^{2}}}\), the chain rule with u = 2x gives
\( \displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1 - (2x)^{2}}} \cdot \frac{d}{dx}(2x) \)
\( \displaystyle = \frac{2}{\sqrt{1 - 4x^{2}}} \)