if we have g: reals^2 --à reals ^2 and f: reals^2 --à reals ^2
And g(x,y) :=((x^2) – (y^2), 2xy) f(u,v) := (u^2) + (v^2)
How do we evaluate (D(f o g))(x,y) by the chain rule.
if we have g: reals^2 --à reals ^2 and f: reals^2 --à reals ^2
And g(x,y) :=((x^2) – (y^2), 2xy) f(u,v) := (u^2) + (v^2)
How do we evaluate (D(f o g))(x,y) by the chain rule.
The "derivative", at a given point, of a function from $\displaystyle R^2$ to $\displaystyle R^2$, at a given point, is a two by two matrix (more correctly, the linear function given by that matrix in the standard basis).
In particular, $\displaystyle Dg= \begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}$
However, you say "f:reals^2 --> reals ^2" but then define "f(u,v) := (u^2) + (v^2)" which is $\displaystyle R^2$ to $\displaystyle R^1$. Assuming you meant to say "f:reals^2-->reals", then Df is a vector (again, more corrrectly, the linear transformation from $\displaystyle R^2$ to $\displaystyle R$ defined by taking the dot product of that vector with a vector v.). Here, that would be [tex]\begin{bmatrix}2u & 2v\end{bmatrix}.
The chain rule says that $\displaystyle D(f\circ g)(x,y)$ is the product of those two matrices:
$\displaystyle \begin{bmatrix}2(x^2- y^2) & 2(2xy)\end{bmatrix}\begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}$