Q84 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $y = \sin(2\sin^{-1} x)$ then $\dfrac{dy}{dx} =$..
- Correct answerA$\dfrac{2 - 4x^2}{\sqrt{1 - x^2}}$
- B$\dfrac{2 + 4x^2}{\sqrt{1 - x^2}}$
- C$\dfrac{2 - 4x^2}{\sqrt{1 + x^2}}$
- D$\dfrac{2 + 4x^2}{\sqrt{1 + x^2}}$
Solution
Given $y = \sin(2\sin^{-1} x)$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \cos(2\sin^{-1} x) \cdot \dfrac{d}{dx}(2\sin^{-1} x)$
$\dfrac{dy}{dx} = \cos(2\sin^{-1} x) \cdot \dfrac{2}{\sqrt{1 - x^2}}$
Using the trigonometric identity $\cos(2\theta) = 1 - 2\sin^2\theta$, we can write:
$\cos(2\sin^{-1} x) = 1 - 2\sin^2(\sin^{-1} x) = 1 - 2x^2$
Substituting this back into the derivative:
$\dfrac{dy}{dx} = (1 - 2x^2) \cdot \dfrac{2}{\sqrt{1 - x^2}}$
$\dfrac{dy}{dx} = \dfrac{2 - 4x^2}{\sqrt{1 - x^2}}$
Differentiating with respect to $x$ using the chain rule:
$\dfrac{dy}{dx} = \cos(2\sin^{-1} x) \cdot \dfrac{d}{dx}(2\sin^{-1} x)$
$\dfrac{dy}{dx} = \cos(2\sin^{-1} x) \cdot \dfrac{2}{\sqrt{1 - x^2}}$
Using the trigonometric identity $\cos(2\theta) = 1 - 2\sin^2\theta$, we can write:
$\cos(2\sin^{-1} x) = 1 - 2\sin^2(\sin^{-1} x) = 1 - 2x^2$
Substituting this back into the derivative:
$\dfrac{dy}{dx} = (1 - 2x^2) \cdot \dfrac{2}{\sqrt{1 - x^2}}$
$\dfrac{dy}{dx} = \dfrac{2 - 4x^2}{\sqrt{1 - x^2}}$
Hence, the correct answer is option: 'A'.