Why is my process of differentiation (trigonometric substitution) not working?

Question

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:

1. It is applicable when we are differentiating inverse trigonometric functions.
2. $x$ 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.
3. 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:

1. Why am I not able to differentiate correctly using the trigonometric substitution method?
2. 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?
3. 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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $sin^{-1}(\sin x)=\neq x \forall x$

Answer 4

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

Answer 5

$$ \sin^{-1}\sin(2\theta) = \left( \{2\theta + 2 \pi k \mid k \in \Bbb{Z}\} \cup \{\pi - 2\theta + 2 \pi k \mid k \in \Bbb{Z}\} \right) \cap [-\pi/2, \pi/2] \text{,} $$ 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.