Proof of Multivariable Chain Rule in higher dimensions

[SpY]]

Homework Statement

Let $$\textbf{F}: \textbf{R}^m \rightarrow \textbf{R}^n$$ and $$\textbf{G}: \textbf{R}^p \rightarrow \textbf{R}^m$$

Prove that $$({\textbf{F} \circ \textbf{G}})'(x) = {\textbf{F}}'(\textbf{G}(\textbf{x})) {\textbf{G}}'(\textbf{x})$$

Homework Equations

Assume the single variable chain rule, that is for
$$f, g: \textbf{R} \rightarrow \textbf{R}$$

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

The Attempt at a Solution

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

Let $$f: \textbf{R}^2 \rightarrow \textbf{R}$$ and $$\textbf{G}: \textbf{R} \rightarrow \textbf{R}^2$$

Then

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

Then by the first mean value theorem,
$$\exists k_1, k_2 \in (0,h)$$ such that

$$G_1 (t+h) - G_1 (t) = h{G_1}'(t+k_1)$$

$$G_2 (t+h) - G_2 (t) = h{G_2}'(t+k_2)$$

Expanding the first two terms previously by substituting $$G_1(t+h)$$
$$f(G_1(t+h), G_2(t+h)) - f(G_1(t), G_2(t+h))$$

$$= f(h{G_1}'(t+k_1) + G_1(t), G_2(t+h))- f(G_1(t), G_2(t+h))$$

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

Where $$p_1 \in (0, h{G_1}'(t+k_1))$$

Similarly for the next two terms substituting $$G_2(t+h)$$
$$f(G_1(t), G_2(t+h)) - f(G_1(t), G_2(t))$$

$$f(G_1(t), h{G_2}'(t+k_2) + G_2(t)) - f(G_1(t), G_2(t))$$

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

Where $$p_2 \in (0, h{G_1}'(t+k_2))$$

Combining this all together and dividing by h:

$$\frac {f(\textbf{G}(t+h)) - f(\textbf{G}(t))}{h}$$

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

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

$${(f \circ \textbf{G})}'(t) =\lim_{h \to 0} \frac {f(\textbf{G}(t+h)) - f(\textbf{G}(t))}{h}$$

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

$$= {f}'(\textbf{G} (t)) { \textbf{G}}'(t)$$

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 $$i \in \{1, ..., n\}, j \in \{1, ..., m\}, k \in \{1, ..., p\}$$.

First we need to have that:

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

Then we can apply the definitions and say:

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

 

[SpY]]

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

Then shouldn't your first line read

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

Because you need to sum the components for $$f$$ otherwise $$f_i$$ 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 $$\frac {\partial} {\partial x_k}$$ not by $$x_j$$ otherwise it goes to $$\frac {\partial} {\partial x_m}$$ because of the sum, or should there be a $$\sum_{k=1}^p$$ 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.