Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another one on chain rule

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


    \frac{\partial}{\partial x} f(2x, 3y)

    First of all, I'm confused by the

    f(2x, 3y)

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

    f(x,y) = cos(xy) - sin(3xy^2}

    and that therefore

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

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

    \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}

    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


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

    for example
    then f(v)= cos(v)=cos(2x)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook