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Derivative of an Integral

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    f(x)= ∫dt/(2 + sin t)
    A and B are the lower and upper limits of the integral, respectively, where A is 0, and B is ln x.

    2. Relevant equations

    3. The attempt at a solution
    g(x)= ln x
    f(g(x))= ∫1/(2+sin t) dt, with a= 0 and b=g(x)
    f'(x)= d/dx ∫1/2+sin x, with a=0, and b=x

    Can't figure out what to do next, and pretty sure what I've done so far is wrong^. Any help is appreciated, thanks so much!
  2. jcsd
  3. Jun 12, 2013 #2


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    Homework Helper

    Here's an easy way to remember it, and you can spend time proving it if you want to :

    $$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x))(b'(x)) - f(a(x))(a'(x))$$
  4. Jun 13, 2013 #3


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    Gold Member

    The original problem was to find f'(x), right? From *, you have f(g(x)). Now apply chain rule and FTOC to find (f(g(x)))' = d (f(g(x)))/dx.
  5. Jun 13, 2013 #4
    Thank you both!

    Using Zondrina's equation, I get:

    {[(1)/(2+sin(ln x))] * (1/x)}

    Is this correct?
  6. Jun 13, 2013 #5


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    Gold Member

    Try what I suggested - what do you get?
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