- #1

- 1,707

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can someone link/show me a formal proof of the multivariable chain rule?

- Thread starter ice109
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- #1

- 1,707

- 5

can someone link/show me a formal proof of the multivariable chain rule?

- #2

- 1,013

- 70

- #3

- 1,707

- 5

what? how is this multavariable?

- #4

lurflurf

Homework Helper

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let x be a vector x=(x1,x2,...,xn)what? how is this multavariable?

- #5

- 406

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There must be a better proof of the chain rule that *that*.

- #6

- 1,707

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yea i thought of that and thought the same thing; that there's gotta be a more rigorous proof.There must be a better proof of the chain rule thanthat.

- #7

lurflurf

Homework Helper

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more rigorous than very rigorousyea i thought of that and thought the same thing; that there's gotta be a more rigorous proof.

interesting

maybe you mean more detailed

that I can provide

Chain rule

let

f:E->F

g:F->G

with E (or a subset) open in F

F (or a subset) open in G

and f differentiable at x

g differentiable at f(x)

then

(g◦f)' exist with

(g◦f)'=[g'◦f][f']

or in more full notation

[g(f(x))]'=[g'(f(x)][f'(x)]

notes on mappings

f:E->F

g:F->G

f':E->L(E,F)

g':F->L(F,G)

(g◦f):E->G

(g◦f)':E->L(E,G)

g'◦f:E->(F,G)

[g'◦f][f']:E->L(E,G)

where L(E,F) is a space of linear mappings from E to F

so all is as it should be

thus derivatives are linear mappings

Δx=dx

Δf:=f(x+dx)-f(x)

df:=f'(x)dx

Δf=df+o(dx) (f differentiable)

informal derivation

dg(f(x))=g'(f(x))df(x)+o(df(x))=g'(f(x))f'(x)dx+o(dx)+f'(x)o(dx)

more formal

let

Δf=df+|dx|rf

rf=(Δf-df)/|dx|

so lim rf=0

now

lim rf=lim rg=0 (f,g differentiable)

Δ[g(f(x))]=g'(f(x))Δf(x)+|Δf|rg(f(x))

=g'(f(x))f'(x)dx+|dx|r(f)+|dx||f'(x)dx/|dx|+rf|rg)

=g'(f(x))f'(x)dx+|dx|{rf+|f'(x)dx/|dx|+rf|rg}

we now need only

lim {r(f)+|f'(x)dx/|dx|+rf(x)|rg(f(x))}=0

which is clear from

rf(x)->0

rg(f(x))->0

and

|f'(x)dx/|dx|+rf|<|f'(x)|+|rf|<∞

ie bounded near x

where |f'(x)| is the norm induced on linear maps by the norm on vectors

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