Simple differentiation question

[Saitama]

Homework Statement

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

Homework Equations

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!

 

[SammyS]

Homework Statement

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

Homework Equations

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]

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

 

[SammyS]

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

 

[Saitama]

\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]

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.

 

[Saitama]

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]

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\,.

 

[Saitama]

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]

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.

 

[Saitama]

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.

 

[SammyS]

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.

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.

 

[Saitama]

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]

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 ...

 

[Saitama]

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]

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 ?

 

[Saitama]

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]

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.

 

[Saitama]

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]

Yeah, just saw those.

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

Hopefully, your overall grade will be more than satisfactory!

 

[Saitama]

Hopefully, your overall grade will be more than satisfactory!

Yes, i hope so.