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Finding a derivative using the FTC Part 1

  1. Sep 1, 2006 #1
    I have a problem which asks me to find the derivative using Part 1 of the Fundamental Theorem of Calculus.

    So I know that the FTC says that if:

    [itex]g(x) = \int_{a}^{h(x)} f(t) dt[/itex] then, [itex]g'(x) = f(h(x)) * h'(x)[/itex]

    I've got what appears to be an easy problem, maybe too easy and because of that I think I'm doing something wrong. Below is the problem and my solution. If someone could just varify if I did it correctly or not, I would greatly appreciate it.

    PROBLEM: [itex]G(x) = \int_{y}^{2} sin(x^2) dx[/itex]

    My Solution:
    [itex]G'(x) = sin(2^2) * 0[/itex]
    [itex] = sin(4) * 0[/itex]
    [itex]= 0[/itex]


    Thanks for taking a look. :smile:
     
  2. jcsd
  3. Sep 1, 2006 #2

    AKG

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    Yes, it's right. Note that G(x) basically depends only on y, so G is a constant w.r.t. x which is why it's no surprise that it's derivative w.r.t. x is 0. It's even clearer if you recognize that the x's in the integrand are just dummy variables so:

    [tex]G(x) = \int _y ^2 \sin (x^2)\, dx[/tex]

    can be rewritten:

    [tex]G(x) = \int _y ^2 \sin (z^2)\, dz[/tex]

    Also, the middle part of: "sin(22)*0 = sin(4)*0 = 0" is unnecessary since regardless of the fact that 22 = 4, the whole thing is something times 0, so it's just 0.
     
  4. Sep 1, 2006 #3

    StatusX

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    Are you sure it isn't G(y)?
     
  5. Sep 2, 2006 #4

    HallsofIvy

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    I'm with StatusX! Any integral of "f(x)dx" with numerical limits is just a constant. The derivative of any constant function is, of course, 0.

    If you are given
    [tex]G(y)= \int_y^2 sin(x^2) dx[/itex]
    and are asked to find
    [tex]\frac{dG}{dy}[/itex]
    or are given
    [tex]G(x)= \int_x^2 sin(t^2)dt[/itex]
    and are asked to find
    [tex]\frac{dG}{dx}[/itex]
    then it would be a non-trivial problem (but still easy).
     
  6. Sep 2, 2006 #5
    Thanks for all of the replies, it's greatly appreciated.
     
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