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.


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 $\arccos$}]$








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.


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 $\arcsin$}]$







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 $\arccos$}]$







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.


  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.

  • 5
    $\begingroup$ Roughly speaking, you must pay very careful attention to domains and ranges when you work with inverse functions. When the domain and range are certain intervals, and when your algebra blithely allows values outside of those intervals, then your computations will be wrong. $\endgroup$
    – Lee Mosher
    Oct 20 at 13:59
  • 3
    $\begingroup$ Consider what happens in your first attempt if we take the convention that if $x = \sin\theta$ then $\sqrt{1-x^2} = \pm\cos\theta.$ Carrying the $\pm$ through, we get $\pm 2/\sqrt{1-x^2},$ two possible values of the derivative for each $x$, of which one is correct for each $x\neq \pm1/\sqrt2.$ The second attempt allows a similar technique with the same result. I don't think the reason for this is quite as obvious as it might seem at first glance, and criteria for selecting the correct sign are still needed, but I find these results curious. $\endgroup$
    – David K
    Oct 20 at 22:21
  • 3
    $\begingroup$ If no-one has commented this yet, it might help your understanding to notice that the expression for the derivative given by the derivative calculator is $-\frac{4x^2-2}{\sqrt{1-4x^2(1-x^2)}\sqrt{1-x^2}}$ which can be simplified to $\frac{2}{\sqrt{1-x^2}} \frac{1-2x^2}{\sqrt{(1-2x^2)^2}}$ which is equal to $\frac{2}{\sqrt{1-x^2}}$ times the sign of $1-2x^2$. $\endgroup$ Oct 20 at 23:10

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.


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 $\arcsin$}]$

“Let $A$ and $(B{\implies}C)$” and “Let $A$ and $B$” and “Let $A$ and $C$” all have different meanings and are not logically equivalent. If you mean “Let $A$ and $B;$ then, since $B{\implies}C, C$”, then, to avoid ambiguity and jarring the reader, just write “Let $A$ and $B;$ then/thus $C$”. Don't use the symbol $\implies$ when you actually mean ‘therefore’.

The entire above chunk can be simply and equivalently written “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 both $(1)$ and $(2)$ are indeed invalid steps. In fact, they are your only missteps thoroughout the two attempts. However, once you perform an invalid step, every subsequent step becomes moot.

    Addendum: This is a rehash of the exact same issue in your previous Question, whose suggested solution's critical error (the invalid step which discarded solutions) was the application of the exact 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.


The mistake is not with the derivative part but the part where you wrote


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

  • 1
    $\begingroup$ Also $\sin^{-1}(\sin2\theta)=2\theta$ is not always true $\endgroup$
    – Asher2211
    Oct 20 at 12:14

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


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


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