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Homework Help: Definition of a derivative in several variables

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello, I'm trying to grasp the definition of a derivative in several variables, that is, to say if it's differentiable at a point.
    My book tells me that a function of two variables if differntiable if:
    [tex]f(a+h,b+k)-f(a,b) = A_1h+A_2k+\sqrt{h^2+k^2}\rho(h,k)[/tex]
    And if [tex] \rho[/tex] goes to zero as (h,k) --> 0. First of all, how did one "come up" with this? It seems a bit arbitrary to me, which I am sure it is not.
    Apart from that, I'm trying to do show that a derivative exists at a point:
    [tex]f(x,y) = sin(x+y) [/tex] at (1,1)

    2. Relevant equations

    3. The attempt at a solution
    To show that the derivative exists:
    [tex]f(1+h,1+k)-f(h,k) = \sin(2+(h+k)) -\sin(2) = 0*h+0*k+\sqrt{h^2+k^2}\rho(h,k)[/tex]
    [tex] \rho(h,k) = \frac{\sin(2+(h+k))}{\sqrt{h^2+k^2}} \text{ if } (h,k) \neq 0 \text{ and } \rho(h,k) = 0 \text{ if } (h,k) = 0 [/tex]
    Then I guess I'm supposed to prove that the limit goes to zero, but how do I do it in this case?
  2. jcsd
  3. Jan 21, 2010 #2


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    Just as the derivative in one variable, of f(x), gives the slope of the tangent line to y= f(x), so the derivative in two variables gives the inclination of a tangent plane to the surface z= f(x,y). Why that exact formula requires a little linear algebra. In general, if [itex]f(x_1, x_2, ..., x_n)[/itex] is a function of n variables, its derivative, at [itex](x_{01}, x_{02}, ..., x_{0n})[/itex] is not a number or a set of numbers but the linear function that best approximates f in some neighborhood of that point. In one variable, any linear function is of the form y= mx+ b. Since b is given by the point, the new information is "m" and we think of that as being the derivative. If [itex]f(x_1, x_2, ..., x_n)[/itex] is a function from [itex]R^n[/itex] to R, then a linear function from [itex]R^n[/itex] to R is of the form [itex]y= a_1x_1+ a_2x_2+ ... + a_nz_n+ b[/itex] which we can think of as [itex]y- b= <a_1, a_2, ..., a_n>\cdot <x_1, x_2, ..., x_n>[/itex], the product being the dot product of two vectors. In that way, we can think of the vector [itex]<a_1, a_2, ..., a_2>[/itex] as being the derivative.

    A more precise definition is this: if [itex]f: R^n\to R^m[/itex], the derivative of f at [itex]x_0= (x_{01}, x_{02}, ..., x_{0n})[/itex] is the linear transformation, L, from [itex]R^n\to R^m[/itex] such that
    [tex]f(x)= f(x_0)+ L(x- x_0)+ \epsilon(x)[/tex]
    for some function [itex]\epsilon[/itex] from [itex]R^n[/itex] to R such that
    [tex]\lim_{x\to x_0} \frac{\epsilon(x)}{|x|}= 0[/tex]
    where |x| is the length of the vector x.

    Of course, that requires that [itex]\epsilon(x)[/itex] go to 0 as x goes to [itex]x_0[/itex] so this is an approximation of f around [itex]x_0[/itex]. The requirement that [itexs\epsilon/|x|[/itex] also go to 0 is essentially the requirement that this be the best linear approximation.

    In the case [itex]R^2\to R[/itex], a "real valued function of two variables", as I said, any linear transformation from [itex]R^2[/itex] to R can be represented as a dot product: [itex]<a, b>\cdot<x, y>= ax+ by[/itex] so the definition above becomes:
    [tex]f(x)= f(x_0, y_0)+ a(x- x_0)+ b(y- y_0)+ \epsilon(x,y)[/tex]
    [tex]lim_{x\to x_0}\frac{\epsilon(x,y)}{|(x-x_0, y- y_0)|}= 0[/itex]
    Of course, in two dimensions that "length" is [itex]\sqrt{(x-x_0)^2+ (y- y_0)^2}[/itex].
    If we take [itex]x- x_0= h[/itex] and [itex]y- y_0= k[/itex] that last says that
    [tex]\frac{\epsilon}{\sqrt{h^2+ k^2}[/tex]
    goes to 0. If we set equal to [itex]\rho(h, k)[/itex] then [itex]\epsilon= \rho(h,k)\sqrt{h^2+ k^2}[/itex] and the result is your formula.

    How did you get "0*h" and "0*k" here? The "A1" and "A2" in your formula are the partial derivatives at the point which, here, are both cos(2), not 0.

  4. Jan 21, 2010 #3
    I'll show you where I got the 0*a from a previous example here. I probably misunderstood something, but just to see what I'm thinking.
    Let's sa we want to show that f(x,y) = xy is differentiable at (1,1).

    [tex]f(1+h,1+k)-f(1,1) = (1+h)(1+k)-1 = h+k+hk = 1*h+1*k+\sqrt{h^2+k^2}*hk/(\sqrt{h^2+k^2} \rightarrow A_1=A_2=1 \and \rho(h,k) = hk/\sqrt{h^2+k^2}[/tex]
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