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Derivative of

  1. Sep 12, 2005 #1
    Using the FTOC and attemting to find the derivative of [tex] g(x) \int^x_0 tan^2 2tdt [/tex] I made [tex] u=tan^2 [/tex]. I know it will wind up being F(x)-F(0) but I'm at a loss with what to do with the 2t. Am I on the right track or have I chosen the wrong u?

    If you could, please guide me and don't spell out the answer.
  2. jcsd
  3. Sep 12, 2005 #2


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    I think the problem is that there are 2 fundamental theorems of calculus. One of them is that:
    where F(x) is an antiderivative of f(x). The other is that
    [tex]\frac{d}{dx}\int_{0}^{x} f(t)dt=f(x)[/tex]
    I think they mean for you to use the latter along with the product rule for differentiation.
  4. Sep 12, 2005 #3


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    Of course, you use the product rule also because of that g(x)
    but the derivative of [tex]\int_0^x f(t)dt[/tex] is just f(x).
  5. Sep 13, 2005 #4
    I've gotten another hint that (tan^2 2t) should have been my u. With that I get (tan2x*tan2x) which I made (sin2x/cos2x)(sin2x/cos2x). Now I make with d/dx g(x)=d/dx[g(u)subx - g(u)sub0] the sub0 drops off because its zero and now I have d/dx[g(u)subx]=f(x) which leaves me with (sin2x/cos2x)(sin2x/cos2x).

    Sorry, got lazy with the la tex.
  6. Sep 13, 2005 #5


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    Why in the world would you change from tan(2x) to [itex]\frac{sin(2x)}{cos(2x)}[/itex]? That just makes it more complicated!

    Representing, for the moment, [itex] \int^x_0 tan^2 2tdt[/itex] as "f(x)", the problem is to differentiate g(x)f(x) and its derivative is, of course, g'(x)f(x)+ g(x)f'(x) (the product rule).

    By the Fundamental Theorem of Calculus (one of them, anyway),
    f'(x)= tan2(2x)- it's as easy as that.

    So your derivative is
    [tex]g'(x)\int^x_0 tan^2 2tdt+ g(x)tan^2(2x)[/tex]
  7. Sep 13, 2005 #6
    ...good grief...I have a long way to go. Thank you.
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