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

Let \( f(x) = \sin^{-1}(2x\sqrt{1 - x^2}) \). 

Let \( u = 2x\sqrt{1 - x^2} \). Then using chain rule: \[ f'(x) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]

Compute \( du/dx \): \[ u = 2x(1 - x^2)^{1/2} \Rightarrow \frac{du}{dx} = 2\sqrt{1 - x^2} + 2x \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}} \]

Compute \( u^2 = 4x^2(1 - x^2) \Rightarrow 1 - u^2 = 1 - 4x^2(1 - x^2) \)

Final derivative: \[ f'(x) = \frac{2x(1 - 2x^2)}{\sqrt{1 - 4x^2(1 - x^2)}} \]

Answer 2

To solve the problem, we need to find the derivative of the function:

$f(x) = \sin^{-1}\left(2x \sqrt{1 - x^2}\right)$

1. Let:
$u = 2x \sqrt{1 - x^2}$

2. Derivative of $f(x)$ using chain rule:
$f'(x) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$

3. Simplify $u$:
$u = 2x (1 - x^2)^{1/2}$

4. Find $\frac{du}{dx}$ using product rule:
$\frac{du}{dx} = 2 \left[ (1 - x^2)^{1/2} + x \cdot \frac{1}{2}(1 - x^2)^{-1/2} (-2x) \right]$
$= 2 \left[ \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}} \right]$
$= 2 \cdot \frac{1 - x^2 - x^2}{\sqrt{1 - x^2}} = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}}$

5. Substitute back into $f'(x)$:
$f'(x) = \frac{1}{\sqrt{1 - (2x \sqrt{1 - x^2})^2}} \cdot \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}}$

6. Simplify denominator:
$1 - (2x \sqrt{1 - x^2})^2 = 1 - 4x^2(1 - x^2) = 1 - 4x^2 + 4x^4 = (1 - 4x^2(1 - x^2))$

7. Final expression:
$f'(x) = \frac{2(1 - 2x^2)}{\sqrt{1 - 4x^2(1 - x^2)} \sqrt{1 - x^2}}$

Final Answer:
The derivative is:
$ { \frac{2x (1 - 2x^2)}{\sqrt{1 - 4x^2 (1 - x^2)} (1 - x^2)} } $ (Option B)