I have a question that says:
Differentiate $\sin^{-1}(4x\sqrt{1-4x^2})$ with respect to $\sqrt{1-4x^2}$, if
(i) $x \in \bigg(-\frac{1}{2\sqrt2},\frac{1}{2\sqrt2} \bigg)$ (ii) $x \in \bigg(\frac{1}{2\sqrt2},\frac{1}{2} \bigg)$ (iii) $x \in \bigg(-\frac{1}{2},-\frac{1}{2\sqrt2} \bigg)$
The answers of the set of questions is: (i)$-\frac{1}{x}$ (ii)$\frac{1}{x}$ (iii)$\frac{1}{x}$
I am not sure how the different intervals are supposed to affect my question. Here's what I tried:
I substituted $2x = \cos \theta$, so y $= \sin^{-1}(4x\sqrt{1-4x^2}) = \sin^{-1}(sin 2\theta)$
I know that $\sin^{-1}(sin 2\theta) = 2\theta$ for $2\theta \in \bigg[\frac{-\pi}{2},\frac{\pi}{2}\bigg]$
So, $\frac{dy}{d\theta} = 2$
and z = $\sqrt{1-4x^2} = \sin \theta$, so I get $\frac{dz}{d \theta} = \cos \theta$
So, I get from here that $\frac{dy}{dz}=\frac{1}{x}$, while the answer says $\frac{-1}{x}$ However, if I had subbed $2x = \sin \theta$, then I would have got the correct answer for this part.
I dont know what I'm supposed to do with the interval of $x$, like, I know that $x \in \bigg(-\frac{1}{2\sqrt2},\frac{1}{2\sqrt2} \bigg)$ so $\cos \theta \in \bigg[\frac{-1}{\sqrt2}, \frac{1}{\sqrt2}\bigg]$, but where does that get me?
Hint:
Using principal values, WLOG $0\le\theta\le\pi\iff0\le2\theta\le2\pi$
and as $\cos\theta$ is decreasing in $\left[0,\pi\right]$
$$\sin^{-1}(\sin2\theta)= \begin{cases}2\theta &\mbox{if } 0\le2\theta\le\dfrac\pi2\iff\cos0\ge \cos\theta\ge\cos\dfrac\pi4\\ \pi-2\theta & \mbox{if } 0\le\pi-2\theta\le\dfrac\pi2\iff\dfrac\pi2\le2\theta\le\pi\\ -2\pi+2\theta & \mbox{if } 0\le-2\pi+2\theta\le\dfrac\pi2\iff\pi\le2\theta\le2\pi+\dfrac\pi2 \end{cases}$$