krcmd1 said:
If P: R2 -> R is defined by p(x,y) = x . y, then
Dp(a,b)(x,y) = bx + ay.
Please tell me in words how to read Dp(a,b)(x,y). Is this a product? a composition of functions? Is this the differential of p(x,y) at (a,b)?
The derivative of a function, from R
2 to R, at a given point in R
2, is a linear transformation from R
2 to R. The form "Dp(a,b)(x,y)" is the linear transformation given by Dp(a,b) applied to the vector (x,y).
It is, in particular, the linear transformation that most closely approximates the functions at that point (all of that can be made precise, of course).
In a given coordinates system, we can always think of a linear transformation from R
2 to R as a "dot product". That is, if, in a given coordinate system, L(x,y)= ax+ by, we can think of the linear transformation as represented by the vector <a, b> so that it's dot product with <x, y> is ax+ by.
Now, for a function from R
2 to R, say f(x,y), that derivative at (a,b), i.e. linear transformation, is represented by the vector
[tex]<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>[/tex]
In this particular case, The partial derivative of xy, with respect to x, is y, and the partial derivative with respect to y is x. At the point (a,b) those are, respectively, b and a. That is, Dp(a,b) is the linear transformation, L, represented by L(x,y)= <b, a>.<x, y>= bx+ ay.
If that's the case, why does the text also state:
If s: R2 -> R is defined by s(x,y) = x + y, then Ds(a,b) = s. (i.e. not "Ds(a,b)(x,y) = ...").
Thank you!
Here, the partial derivatives of x+ y are both 1. Ds(a,b) would be the linear transformation, from R
2 to R, that maps <x,y> to <1, 1>. <x, y>= x+ y. But that is just s itself!
Remember I said, above, that "It is, in particular, the linear transformation that most closely approximates the functions at that point ". Since the given function, s(x,y)= x+ y, is
already linear, Ds= s (or Ds(a,b)= s(a,b)).