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Homework Help: Another one on chain rule

  1. Nov 19, 2005 #1
    Hi,
    here is what I'm trying to do:

    Find

    [tex]
    \frac{\partial}{\partial x} f(2x, 3y)
    [/tex]

    First of all, I'm confused by the

    [tex]
    f(2x, 3y)
    [/tex]

    How does the function look like? I imagine that it is for example

    [tex]
    f(x,y) = cos(xy) - sin(3xy^2}
    [/tex]

    and that therefore

    [tex]
    f(2x, 3y) = cos(6xy) - sin(54xy^2)
    [/tex]

    I'm confused, I don't have any good picture of how it might look in real.
    Anyway, to accomplish the task:

    [tex]
    \frac{\partial}{\partial x} f(2x, 3y) = \frac{\partial f}{\partial x}\frac{d(2x)}{dx} + \frac{\partial f}{\partial y}\frac{d(3y)}{dx} = 2\frac{\partial f}{\partial x}
    [/tex]

    Is it ok? If yes, could you please give an example of this scenario? I mean, how could f(2x, 3y) and f(x,y), respectively, look like so that I could see it with the particular functions?
    Thank you very much.
     
  2. jcsd
  3. Nov 19, 2005 #2
    [tex]f(x,y)[/tex] is just a function mapping of your x and y coordinates.

    As for the derivative, when you take the derivative with respect to x, the y values are treated as if they were constants, same for x when you derive with respect for y.
     
  4. Nov 20, 2005 #3
    So, is my approach ok? And how could the functions look like?

    Thank you.
     
  5. Nov 20, 2005 #4
    Just imagine that

    v(x)=2x, and w(y)=6y

    Then

    f(2x,6y)=f(v(x), w(y)) for arbitrary f.


    for example
    if
    f(x)=cos(x)
    then f(v)= cos(v)=cos(2x)
     
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