If \(y = \sin^{-1}\left(2x\sqrt{1 - x^{2}}\right)\), where \(\displaystyle\frac{1}{\sqrt{2}} \lt x \lt 1\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{2}{\sqrt{1 - x^{2}}}\)
Let \(x = \sin\theta\), with \(\theta = \sin^{-1}x\). The given interval gives \(\frac{\pi}{4} \lt \theta \lt \frac{\pi}{2}\), and \(\sqrt{1 - x^{2}} = \cos\theta\):
\(2x\sqrt{1 - x^{2}} = 2\sin\theta\cos\theta = \sin 2\theta\)
Now \(\frac{\pi}{2} \lt 2\theta \lt \pi\), which is outside the principal range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Since \(\sin(\pi - 2\theta) = \sin 2\theta\) and \(0 \lt \pi - 2\theta \lt \frac{\pi}{2}\):
\(y = \sin^{-1}(\sin 2\theta) = \pi - 2\theta\)
\(y = \pi - 2\sin^{-1}x\)
\( \displaystyle \frac{dy}{dx} = -\frac{2}{\sqrt{1 - x^{2}}} \)
Why the other options are incorrect
Option (1) \(\frac{2}{\sqrt{1 - x^{2}}}\) — Incorrect. This comes from writing \(y = 2\sin^{-1}x\), which holds only for \(-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}\). For example, at x = 0.9 the value y must lie in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), but \(2\sin^{-1}(0.9)\) is about 2.24, more than \(\frac{\pi}{2}\).
Option (3) \(-\frac{1}{\sqrt{1 - x^{2}}}\) — Incorrect. The factor 2 from \(2\theta\) is missing.
Option (4) \(\frac{2x}{\sqrt{1 - x^{2}}}\) — Incorrect. The derivative of \(\pi - 2\sin^{-1}x\) has no factor x, and it is negative on this interval.