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Thm 2.3 (Chain rule):  Suppose $U \subset R^m$ is open and $f: U \rightarrow V \subset \R^m$, $g:  V \rightarrow \R^p$.  Let $h = g \circ f$.  Suppose $f$ is differentiable at $a \in U$ and $g$ is differentiable at $f(a) \in V$.  Then $h$ is differentiable at $a \in U$ and $D(h)_a = D(G)_{f(a)} D(f)_a$.

Let $R_h(\x, \a) = {g( f({\bf x}) ) - g(f({\bf a})) - 
 D(g)_{f(a)} D(f)_a (\x - \a) \over ||{\bf x} - {\bf a}||}$$



Let $\y = f(\x), \b = f(\a)$


$R_g(\y, \b) = {g(\y) - g(\b) - 
 D(g)_{\b} (\y - \b) \over ||{\bf y} - {\bf b}||} $
where $lim_{{\bf x} \rightarrow {\bf a}} R_g(\y, \b) = 0$

$R_f(\x, \a) = {f({\bf x}) - f({\bf a}) - 
 D(f)_a (\x - \a) \over ||{\bf x} - {\bf a}||}$ where $lim_{{\bf x} \rightarrow {\bf a}} R_f(\x, \a) = 0$


$\y - \b = f({\bf x}) - f({\bf a}) = 
 D(f)_a (\x - \a) + ||{\bf x} - {\bf a}||R_f(\x, \a)$ 


$R_g(\y, \b) = {g(f({\bf x}) ) - g(f({\bf a}) ) - 
 D(g)_{\b} [  D(f)_a (\x - \a) + ||{\bf x} - {\bf a}||R_f(\x, \a) ] \over ||{\bf y} - {\bf b}||} $

${||{\bf y} - {\bf b}||R_g(\y, \b) \over ||{\bf x} - {\bf a}||}  = {g(f({\bf x}) ) - g(f({\bf a}) ) - 
 D(g)_{\b}  D(f)_a (\x - \a) -  D(g)_{\b} ||{\bf x} - {\bf a}||R_f(\x, \a)   \over ||{\bf x} - {\bf a}||} $



${||{\bf y} - {\bf b}||R_g(\y, \b) + D(g)_{\b} ||{\bf x} - {\bf a}||R_f(\x, \a)  \over ||{\bf x} - {\bf a}||}  = {g(f({\bf 
x}) ) - g(f({\bf a}) ) - 
 D(g)_{\b}  D(f)_a (\x - \a) \over ||{\bf x} - {\bf a}||} $



$R_h(\x, \a) = {||{\bf y} - {\bf b}||R_g(\y, \b) +  D(g)_{\b} ||{\bf x} - {\bf a}||R_f(\x, \a)  \over ||{\bf x} - {\bf a}||}  
$


$R_h(\x, \a) = {||{f(\x)} - {f(\a)}||R_g(\y, \b) \over ||{\bf x} - {\bf a}||} +  D(g)_{\b}R_f(\x, \a)   $


$R_h(\x, \a) = {|| D(f)_a (\x - \a) + ||{\bf x} - {\bf a}||R_f(\x, \a)||R_g(\y, \b) \over ||{\bf x} - {\bf a}||} +  
D(g)_{\b}R_f(\x, \a)   $



\eject

Cor 2.4:  If $f, g \in C^r$ on $U, V$ respectively, then $h = g \circ f \in \C^r$.  

Proof by induction:

r = 1:  Suppose $f, g \in C^1$  on $U, V$ respectively.  

Then ${\partial f \over \partial x_i}$ exists and is continuous on $U$

Then ${\partial g \over \partial x_i}$ exists and is continuous on $V$.

By Thm 1.3, $f$, $g$ are differentiable on $U, V$ respectively.  

Hence by Thm 1.1, $f$ is continuous.
By Thm 2.3 $h = g \circ f$ is differentiable.  

Thus by Thm 1.1,   ${\partial h \over \partial x_i}$ exist.

By Thm 2.3 it's Jacobian is $D(h)_x = D(g)_{f(x)} D(f)_x$.  

Since ${\partial f_i \over \partial x_j}$ are continuous on $U$ for all 
$i, j$, each entry of \break $D(f)_x = ({\partial f_i \over \partial 
x_j} )$  is continuous

Since  ${\partial g_i \over \partial x_j}$ are continuous on $V$ for all 
$i, j$ and $f$ is continuous, each entry of $D(g)_{f(x)} = 
({\partial g_i \over \partial x_j}|_{f(x)} )$  is continuous.

Since the sums and products of continuous functions are continuous, each entry of $D(h)_a = D(g)_{f(a)} D(f)_a$ is 
continuous.  









\end