If $y=2\sin^{-1}\sqrt{1-x}+\sin^{-1}(2\sqrt{x(1-x)})$ for $0<x<\displaystyle\frac{1}{2}$ then what is the value of $\displaystyle\frac{dy}{dx}$

Question

If $y=2\sin^{-1}\sqrt{1-x}+\sin^{-1}(2\sqrt{x(1-x)})$ for $0<x<\displaystyle\frac{1}{2}$ then $\displaystyle\frac{dy}{dx}$ equals :

A) $\displaystyle\frac{2}{\sqrt{x(1-x)}}$

B) $\displaystyle\sqrt\frac{1-x}{x}$

C) $\displaystyle\frac{-1}{\sqrt{x(1-x)}}$

D) $0$

Substituting $x=\cos^2\theta$, we get $y=2\theta+2\theta=4\theta=4\cos^{-1}\sqrt x$. However derivative is not matching with answer. Answer is given as (D) $0$ .

Answer 1

Let $\displaystyle x=\sin^2\theta\implies 0<\sin^2\theta<\frac12$ where $-\frac\pi4<\theta<\frac\pi4$

$\displaystyle\implies1-x=\cos^2\theta\implies2\sin^{-1}\sqrt{1-x}=2\sin^{-1}\cos\theta=2\left(\frac\pi2-\cos^{-1}\cos\theta\right) =\pi-2\theta$

$\displaystyle2\sqrt{x(1-x)}=2\sin\theta\cos\theta=\sin2\theta\implies\sin^{-1}2\sqrt{x(1-x)}=\sin^{-1}(\sin2\theta)=2\theta$

Reference : Principal values of inverse trigonometric functions

Answer 2

Consider your second term $$\sin^{-1}\Bigl(2\sqrt{x(1-x)}\Bigr)\ .$$ If $0<x<\frac{1}{2}$ and you substitute $x=\cos^2\theta$, then a possible interval of $\theta$ values is $$\frac{\pi}{4}<\theta<\frac{\pi}{2}\ .$$ Then we have $$2\sqrt{x(1-x)}=2\cos\theta\sin\theta=\sin2\theta\ ,$$ with $\frac{\pi}{2}<2\theta<\pi$. Since $2\theta>\frac{\pi}{2}$ we have to be careful with the inverse sine: $$\sin^{-1}(\sin2\theta)=\sin^{-1}(\sin(\pi-2\theta))=\pi-2\theta$$ since $0<\pi-2\theta<\pi/2$. So your complete expression is not $2\theta+2\theta$ but $2\theta+\pi-2\theta$, that is, it is a constant, and the derivative is $0$.