Proof of Multivariable Chain Rule in higher dimensions

[SpY]]

Homework Statement

Let [tex]\textbf{F}: \textbf{R}^m \rightarrow \textbf{R}^n[/tex] and [tex]\textbf{G}: \textbf{R}^p \rightarrow \textbf{R}^m[/tex]

Prove that [tex]({\textbf{F} \circ \textbf{G}})'(x) = {\textbf{F}}'(\textbf{G}(\textbf{x})) {\textbf{G}}'(\textbf{x})[/tex]

Homework Equations

Assume the single variable chain rule, that is for
[tex]f, g: \textbf{R} \rightarrow \textbf{R}[/tex]

[tex]\frac {d(f \circ g)}{dt}(t) = \frac {df}{dt} \big]_{g(t)} \frac {dg}{dt}(t)[/tex]

The Attempt at a Solution

I figured using the single variable result by extending it to [tex]\textbf{R}^2[/tex] first, a sort of subproof which uses the mean value theorem:

Let [tex]f: \textbf{R}^2 \rightarrow \textbf{R}[/tex] and [tex]\textbf{G}: \textbf{R} \rightarrow \textbf{R}^2[/tex]

Then

[tex]f(\textbf{G}(t+h)) - f(\textbf{G}(t)) = f(G_1(t+h), G_2(t+h)) - f(G_1(t), G_2(t+h)) + f(G_1(t), G_2(t+h)) - f(G_1(t), G_2(t))[/tex]
The second and third terms change nothing, I will use them later

Then by the first mean value theorem,
[tex]\exists k_1, k_2 \in (0,h) [/tex] such that

[tex] G_1 (t+h) - G_1 (t) = h{G_1}'(t+k_1) [/tex]

[tex] G_2 (t+h) - G_2 (t) = h{G_2}'(t+k_2) [/tex]

Expanding the first two terms previously by substituting [tex]G_1(t+h)[/tex]
[tex]f(G_1(t+h), G_2(t+h)) - f(G_1(t), G_2(t+h))[/tex]

[tex] = f(h{G_1}'(t+k_1) + G_1(t), G_2(t+h))- f(G_1(t), G_2(t+h))[/tex]

[tex] = h{G_1}'(t+k_1) \frac {\partial df}{\partial dx_1} \big]_{(p_1 + G_1(t), G_2(t+h))} [/tex]

Where [tex]p_1 \in (0, h{G_1}'(t+k_1))[/tex]

Similarly for the next two terms substituting [tex]G_2(t+h)[/tex]
[tex]f(G_1(t), G_2(t+h)) - f(G_1(t), G_2(t))[/tex]

[tex]f(G_1(t), h{G_2}'(t+k_2) + G_2(t)) - f(G_1(t), G_2(t))[/tex]

[tex] = h{G_2}'(t+k_2) \frac {\partial df}{\partial dx_2} \big]_{(G_1(t), p_2 + G_2(t))} [/tex]

Where [tex]p_2 \in (0, h{G_1}'(t+k_2))[/tex]

Combining this all together and dividing by h:

[tex]\frac {f(\textbf{G}(t+h)) - f(\textbf{G}(t))}{h}[/tex]

[tex]= {G_1}'(t+k_1) \frac {\partial df}{\partial dx_1} \big]_{(p_1 + G_1(t), G_2(t+h))} + {G_2}'(t+k_2) \frac {\partial df}{\partial dx_2} \big]_{(G_1(t), p_2 + G_2(t))}[/tex]

Now as [tex]h \rightarrow 0[/tex], [tex]k_1, k_2, p_1, p_2 \rightarrow 0[/tex] since they are contained in intervals up to [tex]h[/tex]. The LHS is now the chain derivative

[tex] {(f \circ \textbf{G})}'(t) =\lim_{h \to 0} \frac {f(\textbf{G}(t+h)) - f(\textbf{G}(t))}{h} [/tex]

[tex] = {G_1}'(t+k_1) \frac {\partial df}{\partial dx_1} \big]_{(p_1 + G_1(t), G_2(t+h))} + {G_2}'(t+k_2) \frac {\partial df}{\partial dx_2} \big]_{(G_1(t), p_2 + G_2(t))}[/tex]

[tex]= {f}'(\textbf{G} (t)) { \textbf{G}}'(t) [/tex]

I've tried generalizing this for any n, but it gets rather long so I'm not sure how to put in concisely. After that, I don't know how to take it to the general proof (any m,n) as required.

Thanks

 

[I like Serena]

Rather than going into all the limit stuff, I think there is an easier way.

Let [tex]i \in \{1, ..., n\}, j \in \{1, ..., m\}, k \in \{1, ..., p\}[/tex].

First we need to have that:

[tex]\frac {\partial} {\partial x_k} f_i(g_1(x), ..., g_m(x)) = \sum_{j=1}^m \frac {\partial} {\partial x_j} f_i(g_1(x), ..., g_m(x)) \cdot \frac {\partial} {\partial x_k} g_j(x))[/tex]

Then we can apply the definitions and say:

[tex](F \circ G)'(x)
= \left( \frac {\partial} {\partial x_k} f_i(g_1(x), ..., g_m(x)) \right)
= \left( \sum_{j=1}^m \frac {\partial} {\partial x_j} f_i(g_1(x), ..., g_m(x)) \cdot \frac {\partial} {\partial x_k} g_j(x)) \right)
= \left( \frac {\partial} {\partial x_j} f_i(g_1(x), ..., g_m(x)) \right) \left( \frac {\partial} {\partial x_k} g_j(x)) \right)
= F'(G(x))G'(x)[/tex]

 

[SpY]]

Hmmm ok so let me get this straight: the [tex]i[/tex] refers to elements in [tex]f[/tex], [tex]j[/tex] to elements in [tex]g[/tex], and [tex]k[/tex] for partial derivatives in [tex]\frac {\partial} {\partial x_k}[/tex]? Where [tex]f: \textbf{R}^n \rightarrow \textbf{R}[/tex] and [tex]g: \textbf{R}^m \rightarrow \textbf{R}[/tex]? (just to be specific on domains here)

Then shouldn't your first line read

[tex]
\sum_{i=1}^n \frac {\partial} {\partial x_k} f_i(g_1(x), ..., g_m(x)) = \sum_{j=1}^m \sum_{i=1}^n \frac {\partial} {\partial x_k} f_i(g_1(x), ..., g_m(x)) \cdot \frac {\partial} {\partial x_k} g_j(x))
[/tex]

Because you need to sum the components for [tex]f[/tex] otherwise [tex]f_i[/tex] is meaningless, then end up with a double sum on the right (over f and g).

Also your first partial derivative on the right should be [tex]\frac {\partial} {\partial x_k}[/tex] not by [tex]x_j[/tex] otherwise it goes to [tex]\frac {\partial} {\partial x_m}[/tex] because of the sum, or should there be a [tex]\sum_{k=1}^p[/tex] somewhere?

I'm having trouble following your last line as well, because you expand the partial derivative using a sum, then just take the sum away keeping the same indices. Throughout you have the variables i, j, k without the sum in front, when you should be summing to n, m, p.

Thanks for the effort though. If a mentor or homework helper could give input it would be appreciated.