Results 1 to 6 of 6

Math Help - How to use chain rule on integrals

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    156

    How to use chain rule on integrals

    Differentiate: ∫ x^2 to 1 ( t^(4) + 1 / t^(2) + 1 )dt

    I dont get how after the step:

    -d/dx ∫ 1 to x^2 ( t^(4) + 1 / t^(2) + 1 )dt

    how you get to: -d/du ∫ 1 to u ( t^(4) + 1 / t^(2) + 1 )dt * du/dx.

    How does using the chain rule affect the upper limit? I also dont really understand the notation, since I am very used to
    f'(g(x))g'(x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,313
    Thanks
    1291
    Quote Originally Posted by SyNtHeSiS View Post
    Differentiate: ∫ x^2 to 1 ( t^(4) + 1 / t^(2) + 1 )dt

    I dont get how after the step:

    -d/dx ∫ 1 to x^2 ( t^(4) + 1 / t^(2) + 1 )dt

    how you get to: -d/du ∫ 1 to u ( t^(4) + 1 / t^(2) + 1 )dt * du/dx.

    How does using the chain rule affect the upper limit? I also dont really understand the notation, since I am very used to
    f'(g(x))g'(x)
    f'(x) is simply another notation for \frac{df}{dx} and f'(g(x)) is the same as \frac{d f(g(x))}{dx} so that f'(g(x))g'(x)= \frac{d f(g(x))}{dg(x)}\frac{dg(x)}{dx}.

    Let F(x)= \int_1^x (t^4+ t^{-2}+ 1)dt. Then F(x^2)= \int_1^{x^2} (t^4+ t^{-2}+ 1)dt and we are seeking -\frac{dF(x^2)}{dx}.

    Let u= x^2 so that F(x^2)= F(u). Now, the chain rule says that \frac{dF}{dx}= \frac{dF}{du}\frac{du}{dx}. Remembering that F(u)= \int_1^u (t^4+ t^{-2}+ 1)dt, that is precisely the formula you are asking about: -\frac{d}{dx}\int_1^{x^2} (t^4+ t^{-2}+ 1)dt=  \left(-\frac{d}{du}\int_1^u (t^4+ t^{-2}+ 1)dt\right)\frac{du}{dx}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    156
    Why do you use the chain rule since the function F(u) isnt a composite (u isnt a function composed in F). Also why isnt the function in terms of t since the function contains a "t" variable and dt?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    t is a 'dummy variable' and often causes confusion. However, if you can just accept the rule that the lower balloon here is the derivative of the upper one with respect to x...



    ... (which is just the FTC) then you can maybe extend to being happy with applying the chain rule. Just in case a picture helps, say



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    Then the FTC with chain rule is...



    In this case...



    Spoiler:




    BTW, correcting small typo here as Lagrange notation's ambiguity on this very point is what H of I was meaning to clarify (I think)...

    Quote Originally Posted by HallsofIvy View Post
    f'(x) is simply another notation for \frac{df}{dx} and f'(g(x)) is the same as g(x)}" alt="\frac{d f(g(x))}{dg(x)}" /> so that f'(g(x))g'(x)= \frac{d f(g(x))}{dg(x)}\frac{dg(x)}{dx}

    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; May 13th 2010 at 10:22 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2010
    Posts
    156
    Howcome you didnt sub in the lower limit and calculate F(x^2) - F(1)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Quote Originally Posted by SyNtHeSiS View Post
    Howcome you didnt sub in the lower limit and calculate F(x^2) - F(1)?
    You could do that (see Differentiation under the integral sign - Wikipedia, the free encyclopedia), but because the derivative of the constant (1) is zero the effect of doing so would be nil. So you can (I think) think of -F(1) as the constant of integration that disappears on the way down. I may try and amend the diagram later to make that clearer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 08:50 PM
  2. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  3. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  4. Replies: 2
    Last Post: December 13th 2007, 05:14 AM
  5. Help with integrals - chain rule
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 23rd 2007, 09:06 AM

Search Tags


/mathhelpforum @mathhelpforum