Why is my process of differentiation (trigonometric substitution) not working? Prologue (you can skip straight to the "Problem" section (bolded) if you want):
First, to show you what way (let's call it trigonometric substitution method) I'm talking about and to show that this way works, I'll describe the tenets and then do a math using that way:
Basic tenets of trigonometric substitution method:

*

*It is applicable when we are differentiating inverse trigonometric functions.

*$x$ should be substituted with a trig ratio that can hold all the possible values of $x$ and that will make differentiation easier. For example, in $\cos^{-1}(\sqrt{\frac{1+x}{2}})$, $-1\leq x\leq1$, so it can be substituted with $\cos\theta$ or $\sin\theta$; substituting with $\sin\theta$ doesn't make our life easier, so we have to substitute with $\cos\theta$. Similarly, in $\tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)$($-1<x\leq1$) and $\sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$$(x\in(\infty,-\infty))$, $x$ has to be substituted with $\cos\theta$ & $\tan\theta$ respectively.

*All of the maths can also be done exclusively using the chain rule. However, the maths might get tedious in that way.

Example
Differentiate with respect to $x$: $\tan^{-1}\frac{4x}{\sqrt{1-4x^2}}.$
Differentiation using trigonometric substitution:
Let, $y=\tan^{-1}\frac{4x}{\sqrt{1-4x^2}}$ and $2x=\cos\theta\implies\theta=\cos^{-1}2x\ [\text{Assuming $\theta$ is within the principal range of $\arccos$}]$
Now,
$$y=\tan^{-1}\frac{4x}{\sqrt{1-4x^2}}$$
$$y=\tan^{-1}\frac{2\cos\theta}{\sqrt{1-\cos^2\theta}}$$
$$y=\tan^{-1}2\cot\theta$$
$$\frac{dy}{dx}=\frac{d}{d(2\cot\theta)}(\tan^{-1}2\cot\theta).\frac{d}{d(\cot\theta)}(2\cot\theta).\frac{d}{d\theta}(\cot\theta).\frac{d}{dx}\theta$$
$$...$$
$$\frac{dy}{dx}=\frac{4}{(12x^2+1)(\sqrt{1-4x^2)}}$$
This is the correct answer. We could've taken $2x=\sin\theta$ as well and the answer would've been the same. We could've done the math exclusively using the chain rule as well.
Problem
Differentiate with respect to $x$: $\sin^{-1}(2x\sqrt{1-x^2}).$
Attempt 1
Let $y=\sin^{-1}(2x\sqrt{1-x^2})$ and $x=\sin\theta\implies\theta=\sin^{-1}x\ [\text{assuming $\theta$ is within the principal range of $\arcsin$}]$
$$y=\sin^{-1}(2x\sqrt{1-x^2})$$
$$y=\sin^{-1}(2\sin\theta\cos\theta)$$
$$y=\sin^{-1}(\sin2\theta)$$
$$y=2\theta\tag{1}$$
$$y=2\sin^{-1}x$$
$$\frac{dy}{dx}=2\frac{1}{\sqrt{1-x^2}}$$
Attempt 2
Let $y=\sin^{-1}(2x\sqrt{1-x^2})$ and $x=\cos\theta\implies\theta=\cos^{-1}x\ [\text{assuming $\theta$ is within the principal range of $\arccos$}]$
$$y=\sin^{-1}(2x\sqrt{1-x^2})$$
$$y=\sin^{-1}(2\sin\theta\cos\theta)$$
$$y=\sin^{-1}(\sin2\theta)$$
$$y=2\theta\tag{2}$$
$$y=2\cos^{-1}x$$
$$\frac{dy}{dx}=-2\frac{1}{\sqrt{1-x^2}}$$
Interestingly enough, we get two different answers using $x=\cos\theta$ & $x=\sin\theta$, which shouldn't have been the case. More importantly, both of the answers are wrong.
Questions:

*

*Why am I not able to differentiate correctly using the trigonometric substitution method?

*In the graph of the correct derivative and the incorrect derivative found using $x=\sin\theta$, there is an overlap between the two from $x=-0.707$ and $x=0.707$. What is the significance of the number $0.707$, and why is the overlap happening?

*In the graph of the correct derivative and the incorrect derivative found using $x=\cos\theta$, there is an overlap between the two from $x=-0.707$ to $x=-1$ in the negative y-axis and from $x=0.707$ to $x=1$ in the positive y-axis. What is the significance of the number $0.707$, and why is the overlap happening?

My observations:
My hunch is that lines $(1)$ & $(2)$ are wrong. However, I don't want to explain my hunch because I fear that it might complicate matters unnecessarily. This might help you in answering the question: it contains the graphs of the original problem, the incorrect derivative found using $x=\sin\theta$, the incorrect derivative found using $x=\cos\theta$ & the correct derivative that can be found by differentiating exclusively using the chain rule.
 A: 
Let $y=\sin^{-1}(2x\sqrt{1-x^2})\,$ and
$x=\sin\theta\implies\theta=\sin^{-1}x\\ [\text{assuming $\theta$ is within the principal range of $\arcsin$}]$

This chunk was confusing to read because you used the symbol $\implies$ when you actually meant ‘therefore’: you wrote “Let $P$ and $[(A$ and $Q){\implies}R]$” but you actually meant “Let $P$ and $A$ and $Q$; thus, $R$”, which isn't equivalent.
Or, equivalently, just write “Let $y=\sin^{-1}(2x\sqrt{1-x^2})$ and $\theta=\sin^{-1}x.$” Note that this implicitly imposes that principal-range restriction.
In general, if we wish to later reverse a substitution, then we do want the substitution to be bijective.

$$y=\sin^{-1}(\sin2\theta);\;\;\therefore y=2\theta\tag{1}$$


$$y=\sin^{-1}(\sin2\theta);\;\;\therefore y=2\theta\tag{2}$$


My hunch is that lines $(1)$ & $(2)$ are wrong.

In each attempt, the specified substitution's principal range allows $2\theta$ to be $\displaystyle\frac34\pi;$ plugging in this value immediately shows that steps $(1)$ and $(2)$ are invalid. They are in fact the only missteps throughout the two attempts.
    Addendum: This is a rehash of the same issue in your previous Question, whose suggested solution's critical error (the invalid step which discarded solutions) was the application of the same false identity $$\arcsin(\sin2\theta) \equiv 2\theta$$ as above!

$$y=\sin^{-1}(2x\sqrt{1-x^2});\;\;\therefore y=\sin^{-1}(2\sin\theta\cos\theta)$$

Note that in both attempts, this step is valid: in attempt 1, $|\cos \theta|\equiv\cos \theta$ in the specified principal range, while in attempt 2, $|\sin \theta|\equiv\sin \theta$ in the specified principal range.
A: The mistake is not with the derivative part but the part where you wrote
$y=\sin^{-1}(2x\sqrt{1-x^2})=2\sin^{-1}x$
You assumed that $\sqrt{1-\sin^2(\theta)}=\cos\theta$ but the correct result is $\sqrt{1-\sin^2(\theta)}=|\cos\theta|$
And as Asher pointed out $sin^{-1}(\sin x)=\neq x \forall x$
A: The equation $\sqrt{1 - \sin^2x} = \cos x$ only holds if $\cos x$ is positive, similarily true for sin-cos switched, so in reality you have $\sqrt{1 - \sin^2x} = \pm \cos x$ depending on the value of $x$. Now $\sin^{-1}$ is an odd function, so in case $\sin x = \theta$ and $\sqrt{1 - \sin^2x} = - \cos x $, we have:
$$y=\sin^{-1}(2x\sqrt{1-x^2}) = -2\theta$$
A: First of all, your use of $\theta$ should be clarified.  This is a new variable that you are introducing, so it is your responsibility to say what it is.  It would be clearer to say "Let $\theta = \sin^{-1} x$"; that specifies what $\theta$ means.  Now you can apply the definition of $\sin^{-1}$ to conclude that $\sin\theta = x$ and $-\pi/2 \le \theta \le \pi/2$.  So the restriction of $\theta$ to the interval $[-\pi/2,\pi/2]$ is not an assumption; it is implied by the definition of $\theta$.
Some answers have questioned your equation $\sqrt{1-\sin^2\theta} = \cos\theta$, but that equation is correct, because $-\pi/2 \le \theta \le \pi/2$.
The mistake is where you go from $y = \sin^{-1}(\sin 2\theta)$ to $y = 2\theta$.  It is not in general true that $\sin^{-1}(\sin \alpha) = \alpha$.  Here is how to fix that step:  $y = \sin^{-1}(\sin 2\theta)$ means $\sin y = \sin 2\theta$ and $-\pi/2 \le y \le \pi/2$.  In other words:  $y$ is the angle in the range $-\pi/2 \le y \le \pi/2$ whose sin is the same as the sin of $2\theta$.  If $-\sqrt{2}/2 \le x \le \sqrt{2}/2$ then $-\pi/4 \le \theta \le \pi/4$, so $-\pi/2 \le 2\theta \le \pi/2$, and in that case it is correct to say that $y = 2\theta$.  But outside of that range, it will not be true that $y = 2\theta$.  If $\sqrt{2}/2 < x \le 1$, then $\pi/4 < \theta \le \pi/2$, so $\pi/2 < 2\theta \le \pi$.  To find $y$, you have to ask:  for what $y$ in the interval $[-\pi/2, \pi/2]$ do we have $\sin y = \sin 2\theta$?  The answer is $y = \pi - 2\theta$.  Similarly, if $-1 \le x < -\sqrt{2}/2$ then you get $y = -\pi-2\theta$.  So the correct formula for $y$ is:
$$
y = 2 \sin^{-1} x \quad \text{if } -\sqrt{2}/2 \le x \le \sqrt{2}/2,
$$
$$
y = \pi - 2 \sin^{-1} x \quad \text{if } \sqrt{2}/2 < x \le 1,
$$
$$
y = -\pi - 2 \sin^{-1} x \quad \text{if } -1 \le x < -\sqrt{2}/2.
$$
Now you can differentiate and get:
$$
\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \quad \text{if } -\pi/2 < x < \pi/2,
$$
$$
\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}} \quad \text{if } -1 < x < -\sqrt{2}/2 \text{ or } \sqrt{2}/2 < x < 1.
$$
The function is not differentiable at $x = \pm \sqrt{2}/2$.  By the way, $\sqrt{2}/2 \approx 0.707$.  That explains the significance of that number.
A: $$ \sin^{-1}\sin(2\theta) = \left( \{2\theta + 2 \pi k \mid k \in \Bbb{Z}\} \cup  \{\pi - 2\theta + 2 \pi k \mid k \in \Bbb{Z}\} \right) \cap [-\pi/2, \pi/2]  \text{,}  $$
which is only $2\theta$ when $2 \theta$ is in the range of arcsine, $[-\pi/2, \pi/2]$.  Otherwise, $2\theta$ is the angle in quadrant 2 or 3 having the same sine, or it's some other coterminal angle of one of these two angles.
