Question

If $ f(x) = \sin^{-1}(2x\sqrt{1 - x^2}) $, then $ f'(x) $ is:

• A. \( \frac{2(1 - 2x^2)}{\sqrt{1 - 4x^2(1 - x^2)}} \)
• B. \( \frac{2x(1 - 2x^2)}{\sqrt{1 - 4x^2(1 - x^2)}} \)
• C. \( \frac{1 - 2x^2}{\sqrt{1 - 4x^2(1 - x^2)}} \)
• D. \( \frac{2x\sqrt{1 - x^2}}{1 - x^2} \)

Hint:

Key Fact: The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \).

Answer 1

The Correct Option is B

Consider the function \( f(x) = \sin^{-1}(2x\sqrt{1 - x^2}) \).

Let \( u = 2x\sqrt{1 - x^2} \). Applying the chain rule, the derivative \( f'(x) \) is given by: \[ f'(x) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]

The derivative of \( u \) with respect to \( x \) is computed as follows: \[ u = 2x(1 - x^2)^{1/2} \Rightarrow \frac{du}{dx} = 2\sqrt{1 - x^2} + 2x \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}} \]

The term \( u^2 \) is calculated as \( u^2 = 4x^2(1 - x^2) \), which implies \( 1 - u^2 = 1 - 4x^2(1 - x^2) \).

The final expression for the derivative \( f'(x) \) is: \[ f'(x) = \frac{2x(1 - 2x^2)}{\sqrt{1 - 4x^2(1 - x^2)}} \]