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 to , 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,
However, you say "f:reals^2 --> reals ^2" but then define "f(u,v) := (u^2) + (v^2)" which is to . Assuming you meant to say "f:reals^2-->reals", then Df is a vector (again, more corrrectly, the linear transformation from to 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 is the product of those two matrices: