Definition of a derivative in several variables

[Gramsci]

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:
f(a+h,b+k)-f(a,b) = A_1h+A_2k+\sqrt{h^2+k^2}\rho(h,k)
And if \rho 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:
f(x,y) = sin(x+y) at (1,1)


2. Relevant equations
-


3. The attempt at a solution
To show that the derivative exists:
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)
and:
\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
Then I guess I'm supposed to prove that the limit goes to zero, but how do I do it in this case?

[HallsofIvy]

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:
f(a+h,b+k)-f(a,b) = A_1h+A_2k+\sqrt{h^2+k^2}\rho(h,k)
And if \rho 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.
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 f(x_1, x_2, ..., x_n) is a function of n variables, its derivative, at (x_{01}, x_{02}, ..., x_{0n}) 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 f(x_1, x_2, ..., x_n) is a function from R^n to R, then a linear function from R^n to R is of the form y= a_1x_1+ a_2x_2+ ... + a_nz_n+ b which we can think of as y- b= <a_1, a_2, ..., a_n>\cdot <x_1, x_2, ..., x_n>, the product being the dot product of two vectors. In that way, we can think of the vector <a_1, a_2, ..., a_2> as being the derivative.

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

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

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

Apart from that, I'm trying to do show that a derivative exists at a point:
[tex]f(x,y) = sin(x+y) at (1,1)


2. Relevant equations
-


3. The attempt at a solution
To show that the derivative exists:
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)
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.

and:
\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
Then I guess I'm supposed to prove that the limit goes to zero, but how do I do it in this case?

[Gramsci]

HallsofIvy:
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).

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}