Method of Differentiation
Inverse Trigonometric Functions
IIT-JEE 1992
Grade 12
Question:
If $y = \sin^{-1}(2x\sqrt{1 - x^2})$ for $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, then $\frac{dy}{dx}$ equals:
(1) $\frac{1}{\sqrt{1 - x^2}}$
(2) $\frac{2}{\sqrt{1 - x^2}}$
(3) $\frac{-2}{\sqrt{1 - x^2}}$
(4) $\frac{2x}{\sqrt{1 - x^2}}$
Step-by-Step Solution
Key Concept: Substitute $x = \sin \theta$ so that $2x\sqrt{1 - x^2} = \sin 2\theta$, reducing $y$ to $2 \sin^{-1} x$ on the given interval before differentiating.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)