TITLE: Ambiguity when finding an Inverse trigonometric derivative
QUESTION [2 upvotes]: 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 $x = \cos\theta$?

REPLY [3 votes]: 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.
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