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

Solution 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.

Solution 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.

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

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