Question

Find the derivative: \[ \frac{d}{dx} \left[ \sin^{-1}\left( 2x \sqrt{1 - x^2} \right) \right] \]

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

Hint:

When dealing with inverse trigonometric functions and composite functions, always use the chain rule and product rule effectively for differentiation.

Answer 1

The Correct Option is C

We are given the function \( \sin^{-1} \left( 2x \sqrt{1 - x^2} \right) \). To find the derivative, we apply the chain rule. First, let: \[ y = \sin^{-1}(u) \quad \text{where} \quad u = 2x \sqrt{1 - x^2}. \] The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is: \[ \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}}. \] Now, we find the derivative of \( u = 2x \sqrt{1 - x^2} \) using the product rule and chain rule: \[ \frac{du}{dx} = 2 \sqrt{1 - x^2} + 2x \cdot \frac{d}{dx} \left( \sqrt{1 - x^2} \right). \] Differentiating \( \sqrt{1 - x^2} \) with respect to \( x \), we get: \[ \frac{d}{dx} \left( \sqrt{1 - x^2} \right) = \frac{-x}{\sqrt{1 - x^2}}. \] Thus, the derivative of \( u \) becomes: \[ \frac{du}{dx} = 2 \sqrt{1 - x^2} + 2x \cdot \frac{-x}{\sqrt{1 - x^2}}. \] Now applying the chain rule: \[ \frac{d}{dx} \left( \sin^{-1} \left( 2x \sqrt{1 - x^2} \right) \right) = \frac{1}{\sqrt{1 - \left( 2x \sqrt{1 - x^2} \right)^2}} \cdot \frac{du}{dx}. \] Simplifying the terms, we find the correct answer is: \[ \frac{2}{\sqrt{1 - x^2}}. \] Thus, the correct answer is option (C).