Understanding Chain Rule: Differentiating P(x, y) and S(x, y)

[krcmd1]

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)? 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!

 

[HallsofIvy]

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 R² to R, at a given point in R², is a linear transformation from R² 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 R² 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 R² 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 R² 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)).

 

[krcmd1]

Thank you.