Question

If \( y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right) \), \( -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \), then \( \frac{dy}{dx} \) is equal to:

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

Hint:

Memorize the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$. It appears frequently in calculus exams and saves significant time over the Chain Rule.

Answer 1

The Correct Option is C

Understanding the Concept: For inverse trigonometric functions, substitution often simplifies the expression before differentiation. The term \( \sqrt{1 - x^2} \) suggests the substitution \( x = \sin \theta \).

Step 1: Substitute \( x = \sin \theta \).
Let \( x = \sin \theta \), then \( \theta = \sin^{-1} x \). The expression becomes: \[ y = \sin^{-1} \left( 2\sin \theta \sqrt{1 - \sin^2 \theta} \right) \] \[ y = \sin^{-1} \left( 2\sin \theta \cos \theta \right) \]

Step 2: Use double angle identity.
Using \( \sin 2\theta = 2\sin \theta \cos \theta \): \[ y = \sin^{-1}(\sin 2\theta) \] Given the range \( -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \), the angle \( 2\theta \) falls within the principal branch \( [-\pi/2, \pi/2] \), so: \[ y = 2\theta \]

Step 3: Substitute back and differentiate.
\[ y = 2\sin^{-1} x \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} \]