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Regular Derivative and A Partial Derivative

  1. Sep 26, 2011 #1
    Can someone please explain to me the difference between a regular derivative and a partial derivative?
     
  2. jcsd
  3. Sep 26, 2011 #2
    I think if you try to define each of the two, you will see the difference. Let us know otherwise.
     
  4. Sep 26, 2011 #3
    If you take both derivatives on a single variable function, they will be the same. What's the definition of the "regular" derivate of a several variables function? Check this

    https://www.physicsforums.com/showthread.php?t=277979
     
  5. Sep 26, 2011 #4
    I didn't mean too be soo abrupt; it is just difficult to answer these questions without more context. If you would tell us some more of what is on your mind?
     
  6. Sep 26, 2011 #5

    Fredrik

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    Suppose that [itex]f:\mathbb R^2\rightarrow\mathbb R[/itex]. The partial derivative of f with respect to the ith variable is a function from [itex]\mathbb R^2\rightarrow\mathbb R[/itex]. I like to denote it by [itex]D_i f[/itex] or [itex]f_{,i}[/itex]. So I would denote the value at (x,y) of the ith partial derivative of f by [itex]D_i f(x,y)[/itex] or [itex]f_{,i}(x,y)[/itex]. I'll stick to the D notation in this post.

    For all values of x and y, [itex]D_1 f(x,y)[/itex] is the value at x, of the derivative of the function [itex]t\mapsto D_1 f(t,y)[/itex]. (Note that this is a function from [itex]\mathbb R\rightarrow\mathbb R[/itex]). In other words, you can define [itex]g:\mathbb R\rightarrow\mathbb R[/itex] by g(t)=f(t,y), and find [itex]D_1 f(x,y)[/itex] by calculating [itex]g'(x)[/itex], because [itex]g'(x)=D_1 f(x,y)[/itex]. We can obviously make a similar comment about partial derivatives with respect to the second variable. So every calculation of the value of a partial derivative at a point in its domain is a calculation of the value of an ordinary derivative at a point in its domain. This is a fact that I don't think is emphasized often enough.

    Example: If you're asked to compute the partial derivative of xy2 with respect to x, it can be interpreted as: Let f be the function defined by f(t)=ty2 for all t. Find f'(x) (i.e. the derivative of f, evaluated at x). If you're asked to compute the partial derivative of xy2 with respect to y, it can be interpreted as: Let g be the function defined by g(t)=xt2 for all t. Find g'(y) (i.e. the derivative of g, evaluated at y).

    [tex]\begin{align}&\frac{\partial}{\partial x}xy^2=(t\mapsto ty^2)'(x)\\ &\frac{\partial}{\partial y}xy^2=(t\mapsto xt^2)'(y)\end{align}[/tex]
    There is really no difference between the expressions [tex]\frac{d}{d x}xy^2[/tex] and [tex]\frac{\partial}{\partial x}xy^2[/tex] for example. The latter is defined to mean [tex]D_1\big((s,t)\mapsto st^2\big)(x,y),[/tex] but this is (by definition of [itex]D_1[/itex]) equal to [tex](s\mapsto sy^2)'(x),[/tex] which is what the former is defined to mean.

    So one valid way of thinking of expressions of the form [tex]\frac{\partial}{\partial x}\big(\text{Something that involves x and at least one more variable}\big)[/tex] is that the partial derivative notation is just telling you which function from [itex]\mathbb R[/itex] into [itex]\mathbb R[/itex] to take an ordinary derivative of, and at what point in the domain to evaluate that derivative.
     
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