proof of chain rule (several variables)
We first consider the case m=1 i.e. G:I→ℝn where I⊂ℝ is a neighbourhood of a point x0∈ℝ and F:U⊂ℝn→ℝ is defined on a neighbourhood U of y0=G(x0) such that G(I)⊂U. We suppose that both G is differentiable 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 notation) we have
F(y0+k)=F(y0)+DF(y0)k+o(|k|). |
Choose any h≠0 such that x0+h∈I 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 h→0 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 value of G′(x0). Thus the product
tends to 0, as needed.
Consider now the general case G:V⊂ℝm→U⊂ℝn.
Given v∈ℝm we are going to compute the directional derivative
∂F∘G∂v(x0)=dF∘gdt(0) |
where g(t)=G(x0+tv) is a function of a single variable t∈ℝ. Thus we fall back to the previous case and we find that
∂F∘G∂v(x0)=DF(G(x0))g′(0).=DF(G(x0))∂G∂v(x0) |
In particular when v=ek is the k-th coordinate vector, we find
g′(0)=DxkF∘G(x0)=DF(G(x0))DxkG(x0)=n∑i=1DyiG(x0)DxkGi(x0) |
which can be compactly written
DF∘G(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 |