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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
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