If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2

Question

If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2 < x < 1/\sqrt 2$, then $\dfrac{dy}{dx}$ equals:

Accepted Answer

A. $\dfrac{2}{\sqrt{1-x^2}}$

Substitute $x = \sin	heta$.
Then $2x\sqrt{1-x^2} = 2\sin	heta\cos	heta = \sin 2	heta$.
So $y = \sin^{-1}(\sin 2	heta) = 2	heta = 2\sin^{-1}x$ (in the given range).
**Hence $dy/dx = 2/\sqrt{1-x^2}$.**

Answer

$\dfrac{1}{\sqrt{1-x^2}}$

Answer

$\dfrac{-2}{\sqrt{1-x^2}}$

Answer

$\dfrac{2}{1+x^2}$

If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2 < x < 1/\sqrt 2$, then $\dfrac{dy}{dx}$ equals:

Keep practising Unit 2: Differentiation and Fundamental Derivative Rules