Ambiguity when finding an Inverse trigonometric derivative

Question

The question I was solving was to find the derivative of $y = \arcsin(2x\sqrt{1-x^2}), -\frac{1}{\sqrt{2} } <x< \frac{1}{\sqrt{2}}$

Substituting $x = \cos\theta \Rightarrow \theta = \arccos (x)$, I got $y = \arcsin(\sin2\theta) = 2\theta$

Differentiating, I got $\frac{-2}{\sqrt{1-x^2}}$

However, substituting $x=\sin\theta$, simplifying and then differentiating, I got $y' = \frac{2}{\sqrt{1-x^2}}$, which is the negative of what I'd gotten earlier. Also, this is the correct answer according to my textbook.

Both substitutions lead to $2x\sqrt{1-x^2}$ turning into $\sin2\theta$, so why is the substitution $x = \sin\theta $ more valid than the substitution $x = \cos\theta$?

@YvesDaoust: anyway, the computation misses details (e.g. $x=\cos\theta$ doesn't imply $\theta=\arccos x$). Cheers! — Bernard, Apr 16 '21 at 10:00
Use https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y — lab bhattacharjee, Apr 16 '21 at 10:01

score 3 · Accepted Answer · answered Apr 16 '21 at 09:44

If $-\frac1{\sqrt2}<x<\frac1{\sqrt2}$, and $x=\cos\theta$, then $\frac\pi4<\theta<\frac{3\pi}4$, and therefore $\frac\pi2<2\theta<\frac{3\pi}2$. But then it is not true that $\arcsin(\sin 2\theta)=2\theta$ (the range of $\arcsin$ is $\left[-\frac\pi2,\frac\pi2\right]$). In fact, $\arcsin(\sin2\theta)=\pi-2\theta$ then.

Ambiguity when finding an Inverse trigonometric derivative

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