1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrating [itex]\frac{\partial f(x,y)}{\partial y(x)}[/itex] with respect to x

  1. Sep 14, 2012 #1
    I don't know how to apply Leibniz's rule for this integrand. I believe there is a substitution to allow me to express this integral differently, but the partial being a function of what we're integrating by is confusing me.

    [itex]\int^{b}_{a} \frac{\partial f(x,y)}{\partial y(x)} dx[/itex]

    Knowns are that [itex][a,b][/itex] is a subset of the domain of [itex]x[/itex] and that [itex]y[/itex] is a function of [itex]x[/itex].

    Can someone please suggest something to get me started, or ask me a question that may get my brain going in the right direction?

    I attempted working with [itex]\int^{b}_{a} \frac{\partial f(x,y)}{\partial x} \frac{dx}{dy}dx[/itex] by expanding the partial according to the chain rule, but I didn't think it helped me.

    Edit: I didn't think the above would help because what exactly does [itex][f(b,y) - f(a,y)]\frac{dx}{dy}[/itex] mean? I don't know how to interpret the factor [itex]\frac{dx}{dy}[/itex].
    Last edited: Sep 14, 2012
  2. jcsd
  3. Sep 14, 2012 #2


    User Avatar
    Science Advisor

    I don't understand your notation. [itex]\partial f/\partial y[/itex] is defined as the derivative of y, treating x as a constant. With that definition, [itex]\partial f/\partial y(x)[/itex] is meaningless. If y is a function of x and x is a constant, then y is a constant, not a variable and we cannot differentiate with respect to it. And the problem you are working on, [itex]\int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx[/itex] does not imply that. You are, I think, using the chain rule incorrectly. You want, of course, "df" so that you can integrate. But the chain rule gives [itex]df= \left(\frac{\partial f}{\partial x}\frac{dx}{dy}+ \frac{\partial f}{\partial y}\right)dy[/itex], not at all what you have- in addition to an additional term, you end with "dx", not "dy".

    The fundamental problem is that your integral, [itex]\int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx[/itex] has an "extra" [itex]\frac{1}{dy}[/itex] that just doesn't make sense in an integral.
  4. Sep 14, 2012 #3
    You can apply it but I don't think it will make much difference. Note:

    [tex]\frac{\partial}{\partial y}\int_a^b f(x,y)dx=\int_a^b \frac{\partial f}{\partial y} dx[/tex]
    Last edited: Sep 14, 2012
  5. Sep 14, 2012 #4
    My original equation was taken from something with messy notation, and I lost some subscripts while working part of the larger question out. So I gave you a poorly framed question to answer -- I'm sorry.

    In my original question, the function [itex]f(x,y)[/itex] is vector valued, and the integrand is with respect to [itex]dx_{1}[[/itex].

    Since I just noticed how sloppy my notation is getting, I'll go back through my problem with more precision before I ask another question.
  6. Sep 14, 2012 #5
    . . . then I want my money back.
  7. Sep 14, 2012 #6
    Don't worry, it's... uh... already in the mail! :uhh:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook