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: Chain Rule:
$y=\sin u$ with $u=2\sin^{-1}x$. Then $\dfrac{dy}{dx}=\cos u\cdot\dfrac{2}{\sqrt{1-x^2}}$.

Step 2: Find cos u:
$\cos u=\cos2\theta=1-2\sin^2\theta=1-2x^2$.

Step 3: Result:
$\dfrac{dy}{dx}=\dfrac{2(1-2x^2)}{\sqrt{1-x^2}}=\dfrac{2-4x^2}{\sqrt{1-x^2}}$. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } \frac{2-4x^2}{\sqrt{1-x^2}}} \]