proof of chain rule (several variables)


We first consider the case m=1 i.e. G:In where I is a neighbourhood of a point x0 and F:Un is defined on a neighbourhood U of y0=G(x0) such that G(I)U. We suppose that both G is differentiableMathworldPlanetmathPlanetmath at the point x0 and F is differentiable in y0. We want to compute the derivative of the compound function H(x)=F(G(x)) at x=x0.

By the definition of derivative (using Landau notationMathworldPlanetmathPlanetmath) we have

F(y0+k)=F(y0)+DF(y0)k+o(|k|).

Choose any h0 such that x0+hI and set k=G(x0+h)-G(x0) to obtain

H(x0+h)-H(x0)h =F(G(x0+h))-F(G(x0))h
=F(G(x0)+k)-F(G(x0))h=F(y0+k)-F(y0)h
=DF(y0)(G(x0+h)-G(x0))+o(|G(x0+h)-G(x0)|)h
=DF(y0)G(x0+h)-G(x0)h+o(|G(x0+h)-G(x0)|)h.

Letting h0 the first term of the sum converges to DF(y0)G(x0) hence we want to prove that the second term converges to 0. Indeed we have

|o(|G(x0+h)-G(x0)|)h|=|o(|G(x0+h)-G(x0)|)|G(x0+h)-G(x0)|||G(x0+h)-G(x0)h|.

By the definition of o() the first fraction tends to 0, while the second fraction tends to the absolute valueMathworldPlanetmathPlanetmathPlanetmath of G(x0). Thus the product tends to 0, as needed.

Consider now the general case G:VmUn. Given vm we are going to compute the directional derivativeMathworldPlanetmath

FGv(x0)=dFgdt(0)

where g(t)=G(x0+tv) is a function of a single variable t. Thus we fall back to the previous case and we find that

FGv(x0)=DF(G(x0))g(0).=DF(G(x0))Gv(x0)

In particular when v=ek is the k-th coordinate vector, we find

g(0)=DxkFG(x0)=DF(G(x0))DxkG(x0)=i=1nDyiG(x0)DxkGi(x0)

which can be compactly written

DFG(x0)=DF(G(x0))DG(x0).
Title proof of chain rule (several variables)
Canonical name ProofOfChainRuleseveralVariables
Date of creation 2013-03-22 16:05:07
Last modified on 2013-03-22 16:05:07
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 6
Author paolini (1187)
Entry type Proof
Classification msc 26B12