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Simple differentiation question

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  • #1
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Homework Statement


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


Homework Equations





The Attempt at a Solution


I started with substituting x=sinθ.
The expression simplifies to [itex]y=\sin^{-1}(\sin(2θ))[/itex] which is equal to [itex]y=2θ[/itex].
Substituting back the value of θ, i get [itex]y=2\sin^{-1}x[/itex].
Therefore [itex]\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}[/itex]
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to [itex]y=2\cos^{-1}θ[/itex], if i differentiate this, i get
[itex]\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}[/itex]
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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  • #2
SammyS
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Homework Statement


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

Homework Equations



The Attempt at a Solution


I started with substituting x=sinθ.
The expression simplifies to [itex]y=\sin^{-1}(\sin(2θ))[/itex] which is equal to [itex]y=2θ[/itex].
Substituting back the value of θ, i get [itex]y=2\sin^{-1}x[/itex].
Therefore [itex]\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}[/itex]
According to the book, this answer is correct.
But if a start by substituting x=cosθ. The expression simplifies to [itex]y=2\cos^{-1}θ[/itex], if i differentiate this, i get
[itex]\frac{dy}{dx}=\frac{-2}{\sqrt{1-x^2}}[/itex]
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!
[itex]\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|[/itex]
 
  • #3
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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.
 
  • #4
SammyS
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... or you could take the sine of both sides and do implicit differentiation.
 
  • #5
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[itex]\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|[/itex]
How does that help? If i take [itex]\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)[/itex], i end up with [itex]\pi+2\cos^{-1}(x)[/itex] 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. :smile:
 
  • #6
SammyS
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How does that help? If i take [itex]\displaystyle \sqrt{\sin^2(\theta)}=-\sin(\theta)[/itex], i end up with [itex]\pi+2\cos^{-1}(x)[/itex] 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.
 
  • #7
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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.
[tex]y=\sin^{-1}(2\cos(\theta)(\sqrt{1-\cos^2(\theta)})[/tex]
[tex]y=\sin^{-1}(2\cos(\theta)(-\sin(\theta)))[/tex]
[tex]y=\sin^{-1}(-sin2(\theta))[/tex]
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.
 
  • #8
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, [itex]-\sin(2\theta)=\sin(-2\theta)\,,[/itex] which looks like will solve your problem.

Furthermore, to do it your way:

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

In other words [itex]-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.[/itex]
 
  • #9
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For one thing, [itex]-\sin(2\theta)=\sin(-2\theta)\,,[/itex] which looks like will solve your problem.
Yes, that solves it. But what about the case when we take [itex]\sqrt{\sin^2(\theta)}=\sin(\theta)[/itex]?

Furthermore, to do it your way:

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

In other words [itex]-\pi/2\le\sin^{-1}(\sin(\pi+2\theta))\le\pi/2\,.[/itex]
How this relation helps out? I know about this but i am still at a loss in understanding how this could help me out. :frown:
 
  • #10
SammyS
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Yes, that solves it. But what about the case when we take [itex]\sqrt{\sin^2(\theta)}=\sin(\theta)[/itex]?
Well, you have this same problem with your first substitution: [itex]x=\sin(\theta)\,.[/itex] It's just that in this case the positive square root works out correctly.

[itex]2x\sqrt{1-x^2}[/itex] becomes

[itex]2\sin(\theta)\sqrt{1-\sin^2(\theta)}[/itex]
[itex]=2\sin(\theta)\sqrt{\cos^2(\theta)}[/itex]

[itex]=2\sin(\theta)(\pm\cos(\theta))[/itex]

[itex]=\pm\sin(2\theta)[/itex]​
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.
 
  • #11
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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:
 
  • #12
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 [itex]\sqrt{1-\cos^2(\theta)} [/itex] which is causing the problem.

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

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

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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  • #13
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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 [itex]\sqrt{1-\cos^2(\theta)} [/itex] which is causing the problem.

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

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

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. :smile:

Good night SammyS.
 
  • #14
SammyS
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I may be doing just small bits at a time.

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

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

but ...
 
  • #15
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but ...
I have got it finally. Here we have [itex]\sin^{-1}(\sin(2\theta))[/itex] where [itex]2\theta[/itex] lies between [itex]\frac{\pi}{2}[/itex] and [itex]\frac{3\pi}{2}[/itex]. 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 [itex]\sin^{-1}(\sin(x))[/itex] is equal to [itex]\pi-2x[/itex].
I have got it now, thank you very much. You saved me. :smile:
 
  • #16
SammyS
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I have got it finally. Here we have [itex]\sin^{-1}(\sin(2\theta))[/itex] where [itex]2\theta[/itex] lies between [itex]\frac{\pi}{2}[/itex] and [itex]\frac{3\pi}{2}[/itex]. 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 [itex]\sin^{-1}(\sin(x))[/itex] is equal to [itex]\pi-2x[/itex].
I have got it now, thank you very much. You saved me. :smile:
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 ?
 
  • #17
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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.
 
  • #18
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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.
 
  • #19
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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. :redface:
 
  • #20
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. :redface:
Hopefully, your overall grade will be more than satisfactory!
 
  • #21
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Hopefully, your overall grade will be more than satisfactory!
Yes, i hope so. :smile:
 

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