Question

The derivative of \(\sin^{-1}(2x\sqrt{1-x^2})\) with respect to \(\sin^{-1}(3x - 4x^3)\) is

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{1}{2}\)
• D. 1

Hint:

Recognizing the triple-angle and double-angle identities simplifies derivative problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Use trigonometric substitutions: Let \(x = \sin\theta\) or \(x = \cos\theta\).

Step 2: Detailed Explanation:
Let \(x = \sin\theta\). Then \(\sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin 2\theta) = 2\theta\) for appropriate range. And \(\sin^{-1}(3x - 4x^3) = \sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta) = 3\theta\). Thus \(u = 2\theta\), \(v = 3\theta\), so \(\frac{du}{dv} = \frac{2}{3}\).

Step 3: Final Answer:
\(\frac{2}{3}\), which corresponds to option (A).