1. Not finding help here? Sign up for a free 30min 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!

Partial Derivatives of an Integral

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the partial derivatives:
    f(x,y)= integral[x,y] cos(t^2)dt, find f_x(x,y) and f_y(x,y)


    2. Relevant equations
    I know from calculus that the derivative of an integral is the function.


    3. The attempt at a solution
    I found that the integral of [x to y] cos(t^2)dt=sin(t^2)= sin(y^2)-sin(x^2)

    and then found the partial derivative of this.
    fx=-2xcos(x^2)
    fy=2ycos(y^2)

    I have seen other procedures and I think I'm wrong. Could someone help?
     
  2. jcsd
  3. Mar 24, 2012 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is wrong. One way to see that is to compute d/dt sin(t^2).

    You need to use the idea you described under "relevant equations". You should start by making your statement of that idea more precise. If g:ℝ→ℝ, then d/dx (what?) =g(x).
     
  4. Mar 24, 2012 #3
    differentiate the integral as a function of x, how is that possible?
    I would need a relation between x or y and t; however, none is shown other than the integral and the function.
     
    Last edited: Mar 24, 2012
  5. Mar 24, 2012 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you can fill in the "what?" in the formula I typed, the rest is easy. t can't have a relationship with either x or y, since t is being integrated over.
     
  6. Mar 24, 2012 #5
    I don't know what to write instead of "what". It could be the integral [x y]: cos(x^2)dx
     
  7. Mar 24, 2012 #6
    Be more general. Fill in the blank with a general expression and then apply it to this problem. How can one, in general, express the antiderivative of [itex]g(x)[/itex]? You already pointed this out in your first post. Now write the general expression.
     
  8. Mar 24, 2012 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Forget about this problem for now, and just explain what you meant when you said that you know that "the derivative of an integral is the function"?
     
  9. Mar 24, 2012 #8
    Ok.
    I have: integral of [ f(x) ], then the derivative of: integral of [f(x)] = f(x).

    or (d/dx)(integral(f(x))=f(x)

    Thank you for your help so far. I can't see things clearly often.
     
  10. Mar 24, 2012 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're still not specific enough. What are the limits of the integration in the formula d/dx[some kind of integral involving f]=f(x)?
     
  11. Mar 24, 2012 #10
    (d/dx)INTEGRAL{f(x)dx }=f(x)
     
  12. Mar 24, 2012 #11

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, you need to specify what interval you're integrating over.

    Edit: OK, I'll just tell you: For all ##a\in\mathbb R##,
    $$\frac{d}{dx}\int_a^x f(t)dt=f(x).$$
     
  13. Mar 24, 2012 #12
    I thought that "t" can't have a relationship with either x or y?
    Why do you relate them?
     
  14. Mar 25, 2012 #13

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I didn't. t clearly doesn't depend on x, and since it makes no difference what variable you use as the dummy variable in the integration (i.e. you can substitute s for t without changing the value of the integral), x doesn't depend on t either.
     
  15. Mar 25, 2012 #14
    Ok, so "t" can be any value from [a , x]?
     
  16. Mar 25, 2012 #15

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I wouldn't say that. Consider the simpler example ##\sum_{k=1}^2a_k##. What value does k have here? The notation is just an abbreviation for ##a_1+a_2##, and k doesn't even appear in that. k only appears in the abbreviation, and can be replaced with any other variable without changing the meaning of the notation. So I wouldn't say that k can be any value in {1,2}. I would say that k isn't even a part of the expression we're working with, and that it's never assigned a value.

    Similarly, ##\int_a^x f(t)dt## is an abbreviation for something that doesn't involve t. This is why these things are called "dummy variables".
     
  17. Mar 25, 2012 #16
    Thank you, It's much more clear now. So, for the original problem. It would be:
    (d/dx) integral from [a to x] f(t)dt =f(x)

    what would "a" be in the original problem?
     
  18. Mar 25, 2012 #17

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You specified that ##f(x,y)=\int_y^x\cos^2 t\, dt##. Not sure I can tell you much more than that because of the forum rule that requires me to only give hints, not solutions. I suggest that you try to compute ##\frac{\partial}{\partial x}f(x,y)##, using that definition of f and the formula I posted. (The f that appears in that formula is of course an arbitrary integrable function).
     
  19. Mar 25, 2012 #18
  20. Mar 25, 2012 #19

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right. Now you have almost solved the problem. Can you figure out how to compute the other partial derivative? (I have to do something else for the next hour or two, so I won't reply as fast).

    You might want to check out the LaTeX guide for the forum. You can also click the quote button next to one of my posts to see how I'm doing the math. (The \, is a small space. It's probably obvious what the other codes I used represent).
     
  21. Mar 25, 2012 #20
    Ok, thank you so far.

    ## \dfrac {d} {dy}\int _{y}^{x}\cos ^{2}\left( t\right) dt=\cos ^{2}\left( y\right) ##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial Derivatives of an Integral
Loading...