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Fundamental Theorem of Calculus Pt 2, multivariable integration?

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    problem in attachment

    2. Relevant equations



    3. The attempt at a solution

    I can get f'(x) as sqrt(1 + (sinx)^2) and derive that to get the second derivative but as far as that I don't really get the concept behind this question

    will y be another function I have to use chain rule with?
    or is it just a variable and I plug in pi/6 normally?
     

    Attached Files:

  2. jcsd
  3. Jan 19, 2012 #2

    SammyS

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    What is [itex]\displaystyle \frac{d}{dy}g(y)\,?[/itex]
     
  4. Jan 19, 2012 #3
    g'(y) = f(y) ?
    g'(y) = f(y)y'

    I think I'm thinking about this too hard...
     
  5. Jan 19, 2012 #4

    SammyS

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    I intended to show the image of your problem in my previous post.

    attachment.php?attachmentid=42829&d=1326954373.jpg

    By the Fund. Thm. of Calc., Pt. 2: g'(y) = f(y). What you use for the independent variable is unimportant, so [itex]g'(x)=f(x)\,.[/itex]

    Then what is [itex]g''(x)\,?[/itex]

    BTW: Welcome to PF !
     
  6. Jan 19, 2012 #5

    HallsofIvy

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    [itex]g(y)= \int_3^y f(x)dx[/itex] is NOT a multiple variable function, it is a function of the single variable y.

    The "fundamental theorem of Calculus" says that g'(y)= f(y). The second derivative of g is simply the first derivative of f.

    [itex]f(u)= \int_0^u\sqrt{1+ t^2}dt[/itex] where [itex]u= sin(x)[/itex]. Now apply both the fundamental theorem of Calculus and the chain rule.
     
  7. Jan 19, 2012 #6
    AH so

    simply f'(pi/6) = g''(pi/6)

    thanks for the help everyone.

    I got something like

    f'(x) = (cosx)sqrt(1 + (sinx)^2)

    plugging in f'(pi/6) = cos(pi/6)sqrt(1 + (sin(pi/6)^2))

    which should be [sqrt(3)/2]*sqrt(5/4) ???
     
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