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Product of function and its derivative

  1. Sep 29, 2011 #1
    Hi there

    Can someone please explain me this step when rewriting the Darcy equation?

    q = -K*h*dh/dx = -0.5*K*d(h^2)/dx

    I don't understand how h*dh/dx changes to 0.5*d(h^2)/dx


    Thank you.

    -Rasmus
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2
    I have tried to seperate variables, integrate on both sides, and then differentiate with regard to x:

    q = -K*h*dh/dx
    <=>
    q dx = -K*h*dh
    <=>
    int q dx = int -K*h*dh
    <=>
    q*x = -0.5*K*h^2
    <=>
    d/dx q*x = d/dx (-0.5*K*h^2)
    <=>
    q = -0.5*K*d(h^2)/dx

    Is that correct?
     
  4. Sep 29, 2011 #3

    Mute

    User Avatar
    Homework Helper

    It's just recognizing the chain rule and reversing it:

    [tex]\frac{dy^2(x)}{dx} = 2y(x)\frac{dy(x)}{dx}[/tex]
     
  5. Sep 29, 2011 #4
    Wow, that took me a long time to understand :-) Thank you.
    Would my suggestion, though much more awkward, still be right?
     
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