# Simple differentiation question

## Homework Statement

Find $\frac{dy}{dx}$.
$$y=\sin^{-1}(2x\sqrt{1-x^2}), \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

## The Attempt at a Solution

I started with substituting x=sinθ.
The expression simplifies to $y=\sin^{-1}(\sin(2θ))$ which is equal to $y=2θ$.
Substituting back the value of θ, i get $y=2\sin^{-1}x$.
Therefore $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to $y=2\cos^{-1}θ$, if i differentiate this, i get
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$
I don't understand why i get these two different answers, i suppose it has to do something with the range of x given in the question but i don't seem to get the point.
Any help is appreciated!

Thanks!

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SammyS
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## Homework Statement

Find $\frac{dy}{dx}$.
$$y=\sin^{-1}(2x\sqrt{1-x^2}), \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$$

## The Attempt at a Solution

I started with substituting x=sinθ.
The expression simplifies to $y=\sin^{-1}(\sin(2θ))$ which is equal to $y=2θ$.
Substituting back the value of θ, i get $y=2\sin^{-1}x$.
Therefore $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to $y=2\cos^{-1}θ$, if i differentiate this, i get
$\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}$
I don't understand why i get these two different answers, i suppose it has to do something with the range of x given in the question but i don't seem to get the point.
Any help is appreciated!

Thanks!
$\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|$

Mark44
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I don't understand why you are using a substitution. This problem requires the use of the chain rule and the product rule.

SammyS
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... or you could take the sine of both sides and do implicit differentiation.

$\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|$
How does that help? If i take $\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)$, i end up with $\pi+2\cos^{-1}(x)$ and i get the wrong answer as before. :(

I don't understand why you are using a substitution. This problem requires the use of the chain rule and the product rule.
Yes, i know about it but don't you see substitution makes it a lot easier. It took some time to write but when i saw the question in my book, i could do it in my mind using substitution.

SammyS
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How does that help? If i take $\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)$, i end up with $\pi+2\cos^{-1}(x)$ and i get the wrong answer as before. :(
Well, you didn't show the steps you took in detail, so I thought perhaps this is where you went wrong.

Well, you didn't show the steps you took in detail, so I thought perhaps this is where you went wrong.

Here are the steps:
Substituting x=cosθ.
$$y=\sin^{-1}(2\cos(\theta)(\sqrt{1-\cos^2(\theta)})$$
$$y=\sin^{-1}(2\cos(\theta)(-\sin(\theta)))$$
$$y=\sin^{-1}(-sin2(\theta))$$
I can rewrite -sin(2θ) as sin(π+2θ).
Therefore, sin-1(sin(π+2θ))=π+2θ=π+2cos-1x.

Please tell me where i am wrong.

SammyS
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I can rewrite -sin(2θ) as sin(π+2θ).
Therefore, sin-1(sin(π+2θ))=π+2θ=π+2cos-1x.

Please tell me where i am wrong.
For one thing, $-\sin(2\theta)=\sin(-2\theta)\,,$ which looks like will solve your problem.

Furthermore, to do it your way:

$\sin^{-1}(\sin(\pi+2\theta))\,,$ is some value between $\pi/2\ \text{ and }\ -\pi/2$

In other words $-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.$

For one thing, $-\sin(2\theta)=\sin(-2\theta)\,,$ which looks like will solve your problem.
Yes, that solves it. But what about the case when we take $\sqrt{\sin^2(\theta)}=\sin(\theta)$?

Furthermore, to do it your way:

$\sin^{-1}(\sin(\pi+2\theta))\,,$ is some value between $\pi/2\ \text{ and }\ -\pi/2$

In other words $-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.$
How this relation helps out? I know about this but i am still at a loss in understanding how this could help me out.

SammyS
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Yes, that solves it. But what about the case when we take $\sqrt{\sin^2(\theta)}=\sin(\theta)$?
Well, you have this same problem with your first substitution: $x=\sin(\theta)\,.$ It's just that in this case the positive square root works out correctly.

$2x\sqrt{1-x^2}$ becomes

$2\sin(\theta)\sqrt{1-\sin^2(\theta)}$
$=2\sin(\theta)\sqrt{\cos^2(\theta)}$

$=2\sin(\theta)(\pm\cos(\theta))$

$=\pm\sin(2\theta)$​
Often with trigonometry expressions, a '±' symbol indicates that you need to choose the correct sign based on context on the situation, rather than in solving algebraic equations where the '±' symbol means that each case is valid.

Often with trigonometry expressions, a '±' symbol indicates that you need to choose the correct sign based on context on the situation, rather than in solving algebraic equations where the '±' symbol means that each case is valid.

So, how would i know which sign to choose here? I have a couple of more problems relating to the same problem where i get two answers. Sorry, if i am acting like a dumb. :uhh:

SammyS
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So, how would i know which sign to choose here? I have a couple of more problems relating to the same problem where i get two answers. Sorry, if i am acting like a dumb. :uhh:
No,not at all.

Looking at all the details certainly helps.

We've only scratched the surface.

It turns out, that it's not the $\sqrt{1-\cos^2(\theta)}$ which is causing the problem.

Yes, $\sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=|\sin(\theta)|\,,$ because if you choose $\displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4}$ to correspond to $\displaystyle -\frac{1}{\sqrt{}2}<x<\frac{1}{\sqrt{2}}\,,$ then $\sin(\theta)>0\,,$ so that $|\sin(\theta)|=\sin(\theta)\ .$

The problem comes from treating $\sin^{-1}(\sin(2\cos^{-1}(x)))$ too casually. After all, $\theta=\cos^{-1}(x)\ .$

But it's late here & I couldn't sleep, so I looked further into this problem, but I really need to try to sleep now.

I'll get back to this later.

Cheers.

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No,not at all.

Looking at all the details certainly helps.

We've only scratched the surface.

It turns out, that it's not the $\sqrt{1-\cos^2(\theta)}$ which is causing the problem.

Yes, $\sqrt{1-\cos^2(\theta)}=\sqrt{\sin^2(\theta)}=|\sin(\theta)|\,,$ because if you choose $\displaystyle \frac{\pi}{4}<\theta<\frac{3\pi}{4}$ to correspond to $\displaystyle -\frac{1}{\sqrt{}2}<x<\frac{1}{\sqrt{2}}\,,$ then $\sin(theta)>0\,,$ so that $|\sin(\theta)|=\sin(\theta)\ .$
Ok, i understand till here. Thanks for the explanation.

The problem comes from treating $\sin^{-1}(\sin(2\cos^{-1}(x)))$ too casually. After all, $\theta=\cos^{-1}(x)\ .$

But it's late here & I couldn't sleep, so I looked further into this problem, but I really need to try to sleep now.

I'll get back to this later.

Cheers.
I can wait for this. Its afternoon here so i have a whole day to prepare for my mathematics exam tomorrow. I am done with all the problems except this one. I guess you will be awake by night(here). Thanks for all the help! I will see if i could myself solve this.

Good night SammyS.

SammyS
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I may be doing just small bits at a time.

While it's always true that $\sin(\sin^{-1}(x))=x\ ,,$ as long as $\sin^{-1}(x)$ is defined, One has to be careful when going the other way. $\sin^{-1}(\sin(\theta))=\theta$ only for $\displaystyle -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\ .$

It looks like you were careful about that in regards to the ends of your interval.

but ...

but ...

I have got it finally. Here we have $\sin^{-1}(\sin(2\theta))$ where $2\theta$ lies between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. I visualized what you said using a graph and i was stunned by seeing how a big fool i am. For this range, the value of function $\sin^{-1}(\sin(x))$ is equal to $\pi-2x$.
I have got it now, thank you very much. You saved me.

SammyS
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I have got it finally. Here we have $\sin^{-1}(\sin(2\theta))$ where $2\theta$ lies between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. I visualized what you said using a graph and i was stunned by seeing how a big fool i am. For this range, the value of function $\sin^{-1}(\sin(x))$ is equal to $\pi-2x$.
I have got it now, thank you very much. You saved me.
Your no fool! It took me a while to see where the problem was arising.

Glad to help! It's good to work with someone willing to think outside the box, and willing to explore a bit on his/her own!

Those "inverse" trig functions can be tricky, what with their limited range & all.

I too used some graphs to get my head around this problem!

Did you happen look at the graph of y if you extend the domain of the problem to -1 ≤ x ≤ 1 ?

Did you happen look at the graph of y if you extend the domain of the problem to -1 ≤ x ≤ 1 ?
Why? I just looked at it on wolframalpha.

SammyS
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Why? I just looked at it on wolframalpha.
Well, there's those sharp cusps at x = ±1/√2 , so, a big discontinuity in the first derivative there.

--- just thought that was interesting.

Well, there's those sharp cusps at x = ±1/√2 , so, a big discontinuity in the first derivative there.

--- just thought that was interesting.

Yeah, just saw those.

Anyways, today was my exam and any question related to this wasn't asked. But i lost marks in matrices.

SammyS
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Yeah, just saw those.

Anyways, today was my exam and any question related to this wasn't asked. But i lost marks in matrices.