Suppose that f is a function of two variables....

Suppose that f is a function of two variables. For small delta x and delta y, we want to find a formula
approximating f(x0 +delta x,y0 +delta y) in terms of the value of f and its partial derivatives at (x0,y0).
Keeping deltax and delta y fixed, consider the function h of one variable:


h(t) = f(x0 +(t)delta x; y0 +(t) delta y):
Note that h(0) = f(x0,y0) and h(1) = f(x0 +delta x; y0 + delta y). Using Taylor formula
h(t) is approximately equal to h(0) + th'(0) +(t^2)/2h''(0)
and so
h(1) is approximately equal to h(0) +h0'(0) +1/2h''(0)

Exercise: Using the Chain rule show that
h'(0) = (delta x)fx(x0, y0) + (delta y)fy(x0, y0)
and

h''(0)=(delta x, delta y)^transpose * M(delta x,delta y),

where M is the matrix
M =

fxx(x0; y0) fxy(x0; y0)
fxy(x0; y0) fyy(x0; y0)


h(t) = f((x0 + t∙Δx), (y0 + t∙Δy))

h'(t) = ∂f/∂x*∂x/∂t + ∂f/∂y*∂y/∂t
h'(t) = fx((x0 + t∙Δx), (y0 + t∙Δy))*Δx + fy((x0 + t∙Δx), (y0 + t∙Δy))*Δy
h'(0) = fx(x0, y0)*Δx + fy(x0, y0)*Δy

h"(t) = Δx*[∂²f/∂x²*∂x/∂t + ∂²f/∂x∂y*∂y/∂t] + Δy*[∂²f/∂y²*∂y/∂t + ∂²f/∂x∂y*∂x/∂t]
h"(t) = Δx*[∂²f/∂x²*Δx + ∂²f/∂x∂y*Δy] + Δy*[∂²f/∂y²*Δy + ∂²f/∂x∂y*Δx]
h"(t) = Δx² ∂²f/∂x² + 2ΔxΔy ∂²f/∂x∂y + Δy² ∂²f/∂y²
h"(0) = Δx² fxx(x0, y0) + 2ΔxΔy fxy(x0, y0) + Δy² fyy(x0, y0)