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: Write the given function.
\[ y=\sin^{-1}(2x\sqrt{1-x^2}) \] We need to find \(\frac{dy}{dx}\) at \(x=0\).

Step 2: Identify inner function.
Let \[ u=2x\sqrt{1-x^2} \] Then, \[ y=\sin^{-1}(u) \]

Step 3: Differentiate using chain rule.
\[ \frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx} \]

Step 4: Differentiate \(u=2x\sqrt{1-x^2}\).
Using product rule: \[ \frac{du}{dx}=2\sqrt{1-x^2}+2x\cdot \frac{d}{dx}(\sqrt{1-x^2}) \] Now, \[ \frac{d}{dx}(\sqrt{1-x^2})=\frac{-x}{\sqrt{1-x^2}} \] So, \[ \frac{du}{dx}=2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}} \]

Step 5: Evaluate at \(x=0\).
At \(x=0\), \[ u=0 \] So, \[ \frac{1}{\sqrt{1-u^2}}=1 \] Also, \[ \frac{du}{dx}=2(1)-0=2 \]

Step 6: Substitute values.
\[ \frac{dy}{dx}=1\cdot 2=2 \]

Step 7: Final answer.
\[ \boxed{2} \] which matches option \((3)\).