Question

If \(y=\sin^{-1}(2x\sqrt{1-x^2})\), then \(\frac{dy}{dx}\) at \(x=0\) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(\frac{\sqrt{3}}{2}\)
• E. \(-1\)

Hint:

In inverse trigonometric functions, always use chain rule carefully. Evaluating at a specific point often simplifies the expression significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves differentiating an inverse trigonometric function. The expression inside the \( \sin^{-1} \) function has a specific form that suggests a trigonometric substitution to simplify the function before differentiation.
Step 2: Key Formula or Approach:
The trigonometric identity for the sine of a double angle is \( \sin(2\theta) = 2\sin\theta\cos\theta \).
The expression \( 2x\sqrt{1-x^2} \) resembles this identity. We can use the substitution \( x = \sin\theta \). This implies \( \theta = \sin^{-1}x \) and \( \sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta \).
After simplifying, we differentiate the resulting function and evaluate it at \( x=0 \).
Step 3: Detailed Explanation:
Let's apply the substitution \( x = \sin\theta \). The function \( y \) becomes:
\[ y = \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta}) \] \[ y = \sin^{-1}(2\sin\theta\cos\theta) \] Using the double angle identity:
\[ y = \sin^{-1}(\sin(2\theta)) \] For the appropriate range of \( \theta \), this simplifies to:
\[ y = 2\theta \] Now, substitute back \( \theta = \sin^{-1}x \):
\[ y = 2\sin^{-1}x \] This simplified form is much easier to differentiate. The derivative of \( \sin^{-1}x \) is \( \frac{1}{\sqrt{1-x^2}} \).
\[ \frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1}x) = 2 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}} \] Finally, we evaluate this derivative at \( x=0 \):
\[ \left. \frac{dy}{dx} \right|_{x=0} = \frac{2}{\sqrt{1-0^2}} = \frac{2}{\sqrt{1}} = 2 \] Step 4: Final Answer:
The value of \( \frac{dy}{dx} \) at \( x=0 \) is 2. This corresponds to option (C).