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Derivatives of integrals and inverse functions

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of:

    [itex]1. f(x)=arccos(5x^3)[/itex]

    [itex]2. f(x)=∫cos(5x)sin(5t)dt[/itex] when the integral is from 0 to x


    2. Relevant equations

    Chain rule, dy/dx=dy/du*du/dx

    3. The attempt at a solution

    For the first one, I can just take 5x^3 as u and then apply the chain rule to get

    [itex]-15x^2/\sqrt{1-5x^3}[/itex]

    but maple TA keeps telling me that it's incorrect...

    As for the second one, I'm not sure how to approach something like that when x is actually in the integral, as in the cos(5x). I mean, when I'm applying the chain rule...do I have to include cos in my choice of u?
     
  2. jcsd
  3. Sep 15, 2012 #2
    Check the derivative of arccos again.
    [tex]\frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}[/tex]

    As for the second one, when you have something like this:
    [tex]\int_{g(x)}^{h(x)} f(t)dt[/tex]
    It's derivative is equal to
    [tex]f(g(x)) \cdot g'(x)-f(h(x)) \cdot h'(x)[/tex]
     
  4. Sep 15, 2012 #3
    Ah, so the derivative of the first one should be

    [itex]-15x^2/\sqrt{1-25x^6}[/itex]?
     
  5. Sep 15, 2012 #4
    Yep! :smile:
     
  6. Sep 16, 2012 #5
    Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? i.e. I only have to multiply by the derivative of x?

    I know what to do if I have something like

    [itex]d/dx∫cos(5t)sin(5t)dt[/itex], where the integral is from 0 to x. I just differentiate by basically putting the x into the ts to get

    [itex]cos(5x)sin(5x)[/itex]

    But when the cos5t in the integral becomes cos5x, do I just do this?

    [itex]cos(5x)sin(5x)*(d/dx)x=cos(5x)sin(5x)
     
  7. Sep 16, 2012 #6
    Yep, just take out the cos(5x) from the integral.
    You are left with
    [tex]f(x)=cos(5x) \int_{0}^{x} sin(5t)dt[/tex]
    You can either differentiate by the chain rule or better, solve the integral, put the limits and differentiate.
     
  8. Sep 16, 2012 #7
    Great, I finally got it, thanks again :)
     
  9. Sep 16, 2012 #8

    SammyS

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    Don't forget, [itex]\displaystyle \int_{0}^{x} \sin(5t)dt[/itex] is a function of x, so you have to take that into account when you differentiate [itex]\displaystyle \cos(5x)\int_{0}^{x} \sin(5t)dt\ .[/itex]
     
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