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What is a function, when you consider the partial/total derivative?

  1. Feb 27, 2014 #1
    Example, if:
    [itex]z=xy[/itex]
    [itex]y=x[/itex]
    then:
    the partial derivative [itex]\frac{\partial z}{\partial x} = y[/itex], treating [itex]z[/itex] as a function of two variables [itex]z(x, y) = xy[/itex]
    the total derivative [itex]\frac{dz}{dx} = 2x[/itex], treating [itex]z[/itex] as a function of one variable, [itex]z(x) =z y(x)[/itex].

    So there doesn't seem to be a way to define the partial or total derivative based only on one function [itex]z[/itex] defined as a set of ordered pairs plus domain/codomain. So how do you define a function in such a way that you can define the partial and total derivative?


    This may not be clear. If you don't see the problem, consider this question:
    For [itex]f : \mathbb{R} \times \mathbb{R} -> \mathbb{R}[/itex], what's the total derivative of [itex]f[/itex] with respect to its first parameter, and how does this differ from the partial derivative of [itex]f[/itex] with respect to its first parameter?

    How many parameters does the operator "total derivative" take and what are they? There's at least the function and the variable the total derivative is taken with respect to, but those two parameters don't seem enough, without changing the definition of function. How many parameters does the operator "partial derivative" take and what are they?

    To handle the distinction between the total derivative and partial derivative, in calculus you also need some idea of the relationships between the variables used in the function definition. These relationships are not encoded in the usual definition of a function.
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2

    tiny-tim

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    hi mXSCNT! :smile:
    correct :smile: … ∂/∂x depends on what the other variable(s) are, since you have to calculate it while keeping all the other variables constant

    so eg ∂/∂x f(x,y,z) = d/dx fy,z(x), where fy,z(x) = f(x,y,z) :wink:

    (this is not a standard notation)
     
  4. Feb 28, 2014 #3

    pasmith

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    You don't.

    Given a differentiable [itex]f: \mathbb{R}^2 \to \mathbb{R}[/itex] you can define these partial derivatives (I'm avoiding the Liebnitz notation because it tends to obscure matters in rigorous analysis):
    [tex]
    f_1: \mathbb{R}^2 \to \mathbb{R} : (x,y) \mapsto \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}h \\
    f_2: \mathbb{R}^2 \to \mathbb{R} : (x,y) \mapsto \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}h
    [/tex]
    These respectively are the partial derivatives of [itex]f[/itex] with respect to its first and second arguments.

    Given a differentiable [itex]f: \mathbb{R}^2 \to \mathbb{R}[/itex] and a differentiable [itex]g: \mathbb{R} \to \mathbb{R}[/itex], you can then define an entirely new function
    [tex]
    h: \mathbb{R} \to \mathbb{R}: x \mapsto f(x,g(x))[/tex]
    whose derivative can be shown to be
    [tex]
    h' : x \mapsto f_1(x, g(x)) + f_2(x, g(x))g'(x)[/tex]
    Note that [itex]f \neq h[/itex] since the domains are different: the domain of the first is [itex]\mathbb{R}^2[/itex] and the domain of the second is [itex]\mathbb{R}[/itex]. In these circumstances, writing
    [tex]
    \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}[/tex]
    is a notational abuse (one of the many to which the Liebnitz notation lends itself).
     
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