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Simple differentiation question

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data
    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]


    2. Relevant equations



    3. 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!
     
    Last edited: Sep 24, 2012
  2. jcsd
  3. Sep 24, 2012 #2

    SammyS

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    [itex]\displaystyle \sqrt{\sin^2(\theta)}=|\sin(\theta)|[/itex]
     
  4. Sep 24, 2012 #3

    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.
     
  5. Sep 24, 2012 #4

    SammyS

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    ... or you could take the sine of both sides and do implicit differentiation.
     
  6. Sep 24, 2012 #5
    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. :(

    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:
     
  7. Sep 24, 2012 #6

    SammyS

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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.
     
  8. Sep 24, 2012 #7
    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.
     
  9. Sep 24, 2012 #8

    SammyS

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    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]
     
  10. Sep 24, 2012 #9
    Yes, that solves it. But what about the case when we take [itex]\sqrt{\sin^2(\theta)}=\sin(\theta)[/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:
     
  11. Sep 25, 2012 #10

    SammyS

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    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.
     
  12. Sep 25, 2012 #11
    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:
     
  13. Sep 25, 2012 #12

    SammyS

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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]

    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.
     
    Last edited: Sep 25, 2012
  14. Sep 25, 2012 #13
    Ok, i understand till here. Thanks for the explanation. :smile:

    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.
     
  15. Sep 25, 2012 #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 ...
     
  16. Sep 25, 2012 #15
    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:
     
  17. Sep 25, 2012 #16

    SammyS

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    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 ?
     
  18. Sep 25, 2012 #17
    Why? I just looked at it on wolframalpha.
     
  19. Sep 25, 2012 #18

    SammyS

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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.
     
  20. Sep 26, 2012 #19
    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:
     
  21. Sep 26, 2012 #20

    SammyS

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    Hopefully, your overall grade will be more than satisfactory!
     
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