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Fundamental theorem of calculus

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of cost/t (dt) evaluated from 3 to the sqrt of x

    3. The attempt at a solution
    using the fundamental theorem of calculus, I have [cos(sqrt x)]/(sqrtx)
    I know I have to use the chain rule but I don't know how to go about it from here. Any tips?
  2. jcsd
  3. Oct 31, 2007 #2


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    You just have to differentiate your function of x. Start with the derivative of cos(sqrt(x)). What is it? If you can get that using the chain rule the just use the quotient rule to do the whole thing.
  4. Nov 1, 2007 #3

    Gib Z

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    Are you trying to find the integral or the derivative? It says derivative, but "evaluated from 3 to the sqrt of x" implies integral :( Also, using FTC also implies you are finding the integral...
  5. Nov 1, 2007 #4


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    I think this is the question, basically.

    Evaluate [tex]\frac{d}{dx}\int_3^{\sqrt{x}}\frac{\cos{t}}{t}dt[/tex] and that's a trivial application of FTC.

    Try making the substitution [tex]u= \sqrt{x}[/tex], working out [tex]\frac{d}{du}\int_c^{u}\frac{\cos{t}}{t}dt[/tex], then using the chain rule [tex]\frac{dI}{dx} = (\frac{dI}{du})(\frac{du}{dx})[/tex] (c is an arbitrary constant,the lower bound does not really matter).
    Last edited: Nov 1, 2007
  6. Nov 1, 2007 #5


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    That makes no sense! You mean "find the derivative of the integral of cost/t dt evaluated from 3 to the sqrt(x)[/itex]

    Perhaps it would make sense if you wrote f(u)= int{3 to u} cos(t)/t dt and u(x)= sqrt(x). Now apply the chain rule to those two functions: df(x)/dx= df(u)/du du/dx. both of those should be easy now.
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