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Simple question about partial derivatives

  1. Aug 21, 2011 #1
    I have come to a bit of a misunderstanding with partial derivatives. I will try to illustrate my problem. Say we have a function f(x, y(x), y'(x)) where y'(x)=dy/dx. Now suppose that f does not explicitly depend on x. My physics book says at this point that ∂f/∂x=0, even though y(x) and y'(x) may depend on x.
    Suppose f=y2
    Then ∂f/∂x=0
    but if we have y(x)=x, then we can write f as:
    f=x2 and we have
    ∂f/∂x=2x
    How can we have two different answers for the same derivative by simply rewriting the function in a different way?
    I apologize in advance if the answer is obvious and I am being a bit annoying by asking. But if you do have a helpful comment to post, I would greatly appreciate it!
    -Alex
     
  2. jcsd
  3. Aug 21, 2011 #2

    mathman

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    In the first case, ∂f/∂y = 2y. Note that (I'll ignore y' since it is irrelevant):
    df/dx = ∂f/∂x + (∂f/∂y)(dy/dx). Therefore the final result is the same in both cases.
     
  4. Aug 22, 2011 #3

    HallsofIvy

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    The partial derivative of f "with respect to y" assumes y is independent of x or any other variables. If, in fact, f is a function of both x and y, and y is a function of x, then
    [tex]\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}[/tex]

    In your example, with [itex]f(x,y)= y^2[/itex] and y= x,
    [tex]\frac{\partial f}{\partial x}= 0[/tex]
    [tex]\frac{\partial f}{\partial y}= 2y[/tex]
    and
    [tex]\frac{df}{dx}= 0+ (2y)(1)= 2x[/tex].
     
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