If \(y = \sin^{-1}\left(3x - 4x^{3}\right)\), where \(-\displaystyle\frac{1}{2} \lt x \lt \frac{1}{2}\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3}{\sqrt{1 - x^{2}}}\)
Let \(x = \sin\theta\), with \(\theta = \sin^{-1}x\). The given interval gives \(-\frac{\pi}{6} \lt \theta \lt \frac{\pi}{6}\). By the triple-angle formula:
\(3x - 4x^{3} = 3\sin\theta - 4\sin^{3}\theta = \sin 3\theta\)
Since \(-\frac{\pi}{2} \lt 3\theta \lt \frac{\pi}{2}\), which is inside the principal range of \(\sin^{-1}\):
\(y = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x\)
\( \displaystyle \frac{dy}{dx} = \frac{3}{\sqrt{1 - x^{2}}} \)