Question

If \(y = sin(2sin^{-1}x)\) then \(\frac{dy}{dx} =\)..

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

Hint:

Put x = sin t, so y = sin 2t, and differentiate with the chain rule.

Answer 1

The Correct Option is A

Step 1: Substitute:
Let \(\theta=\sin^{-1}x\), so \(x=\sin\theta\). Then \(y=\sin2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}\).

Step 2: Differentiate With the Product Rule:
\[ \frac{dy}{dx}=2\sqrt{1-x^2}+2x\cdot\frac{-x}{\sqrt{1-x^2}} \]

Step 3: Combine:
\[ \frac{dy}{dx}=\frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}}=\frac{2-4x^2}{\sqrt{1-x^2}} \]

Step 4: Check the Options:
Options (C) and (D) have \(\sqrt{1+x^2}\) in the denominator, which cannot arise from derivatives of \(\sin^{-1}\). Option (B) has \(+4x^2\), but the second term is subtracted. So (A) is correct.

Final Answer:
\(\dfrac{dy}{dx}=\dfrac{2-4x^2}{\sqrt{1-x^2}}\), option (A). \[ \boxed{\text{(A) } \frac{2-4x^2}{\sqrt{1-x^2}}} \]