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

Does a derivative distribute?

  1. May 24, 2010 #1
    I'm trying to simplify this. I have two functions, f0(z), and f'(t, x, y, z). One is the 'base' value that varies only with height, the other is the small 'perturbation' value that varies with all four variables.

    I am substituting these into an equation that calls for ∂(f)/∂(x)

    Do I write this as ∂ (f0 + f' ) / ∂(x) ?

    And if so, how can I simplify it? Since f0 is not a function of x, can I just cross it out completely? Is it equal to ∂(f0) / ∂(x) + ∂(f') / ∂(x) ?
  2. jcsd
  3. May 24, 2010 #2
    What's "f"?
    Oh, f is the sum of the two things.

    And yes, you're right.
    Derivative of a sum equals the sum of derivatives.
    And yes, ∂(f0) / ∂(x)=0.
  4. May 24, 2010 #3
    In this case it is density...

    By the way I'd also like to know the general answer here. That is, if the f0 term were a function of x as well.
  5. May 24, 2010 #4
    Sorry, I didn't pick up that f:=f0 + f '
    somehow. lol
  6. May 24, 2010 #5


    User Avatar
    Homework Helper

    Yes. The derivative is a linear operation, so it distributes. You can see this from the definition. If a function f depends on coordinates x_0,x_1,...,x_n, then the partial derivative of f with respect to one of the x's, x_i for generality, is

    [tex]\lim_{h \rightarrow 0} \frac{f(x_0,x_1,\dots,x_i+h,\dots,x_n) - f(x_0,x_1,\dots,x_i,\dots,x_n)}{h} = \frac{\partial f(x_0,x_1,\dots,x_i,\dots,x_n)}{\partial x_i}[/tex]

    If you plug in f(x_0,x_1,...,x_n) = f_0(x_0,x_1,...,x_n) + f_1(x_0,x_1,...,x_n), you can rearrange that so that you get the sum of the derivatives. Assuming that f_0 depends only on x_j (for j not equal to i) causes that term to drop out.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook