If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2 < x < 1/\sqrt 2$, then $\dfrac{dy}{dx}$ equals:
A$\dfrac{2}{\sqrt{1-x^2}}$
B$\dfrac{1}{\sqrt{1-x^2}}$
C$\dfrac{-2}{\sqrt{1-x^2}}$
D$\dfrac{2}{1+x^2}$
Answer & Solution
Correct answer: A. $\dfrac{2}{\sqrt{1-x^2}}$
Substitute $x = \sin\theta$. Then $2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta$. So $y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$ (in the given range). Hence $dy/dx = 2/\sqrt{1-x^2}$.
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