7. The Chain Rule

本节主要说明链式法则，即多维空间中，可导函数的复合也是可导的。Theorem 7.1是本节的主要定理。Corollary 7.2说明$C^r$类函数的复合也是$C^r$类。Theorem 7.3是多维空间的均值定理。Theorem 7.4说明多维空间下反函数定理也成立，但是反函数的导数变为原函数导数的逆矩阵，由此可知，反函数可导需要原函数导数矩阵非奇异。

####Exercises
Exercise 1. Let $f:{\mathbf{R}}^3\to {\mathbf{R}}^2$ satisfy the condition $f(0)=(1,2)$ and
$$Df(0)=\left[\begin{array}{ccc}1& 2& 3\\ 0& 0& 1\end{array}\right]$$
Let $g:{\mathbf{R}}^2\to {\mathbf{R}}^2$ be defined by the equation
$$g(x,y)=(x+2y+1,3xy)$$
Find $D(g\circ f)(0)$.
Solution: We first have
$$Dg(x,y)=\left[\begin{array}{cc}1& 2\\ 3y& 3x\end{array}\right],Dg(1,2)=\left[\begin{array}{cc}1& 2\\ 6& 3\end{array}\right]$$
then by the chain rule
$$\begin{array}{rl}D(g\circ f)(0)& =Dg(f(0))\cdot Df(0)=Dg(1,2)\cdot Df(0)\\ & =\left[\begin{array}{cc}1& 2\\ 6& 3\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 5\\ 6& 12& 21\end{array}\right]\end{array}$$

Exercise 2. Let $f:{\mathbf{R}}^2\to {\mathbf{R}}^3$ and $g:{\mathbf{R}}^3\to {\mathbf{R}}^2$ be given by the equations
$$f(x)=(e^{2x_1+x_2},3x_2-\cos x_1,x_1^2+x_2+2)$$
$$g(y)=(3y_1+2y_2+y_3^2,y_1^2-y_3+1)$$
( a ) If $F(x)=g(f(x))$, find $DF(0)$.
( b ) If $G(y)=f(g(y))$, find $DG(0)$.
Solution: We first have $f(0,0)=(1,-1,2),g(0,0,0)=(0,1)$ and
$$Df(x_1,x_2)=\left[\begin{array}{cc}2e^{2x_1+x_2}& e^{2x_1+x_2}\\ \sin x_1& 3\\ 2x_1& 1\end{array}\right],Dg(y_1,y_2,y_3)=\left[\begin{array}{ccc}3& 2& 2y_3\\ 2y_1& 0& -1\end{array}\right]$$
( a ) By the chain rule we have
$$\begin{array}{rl}DF(0)& =Dg(f(0))\cdot Df(0)=Dg(1,-1,2)\cdot Df(0,0)\\ & =\left[\begin{array}{ccc}3& 2& 4\\ 2& 0& -1\end{array}\right]\left[\begin{array}{cc}2& 1\\ 0& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}6& 13\\ 4& 1\end{array}\right]\end{array}$$
( b ) By the chain rule we have
$$\begin{array}{rl}DG(0)& =Df(g(0))\cdot Dg(0)=Df(0,1)\cdot Dg(0,0,0)\\ & =\left[\begin{array}{cc}2e& e\\ 0& 3\\ 0& 1\end{array}\right]\left[\begin{array}{ccc}3& 2& 0\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{ccc}6e& 4e& -e\\ 0& 0& -3\\ 0& 0& -1\end{array}\right]\end{array}$$

Exercise 3. Let $f:{\mathbf{R}}^3\to \mathbf{R}$ and $g:{\mathbf{R}}^2\to \mathbf{R}$ be differentiable. Let $F:{\mathbf{R}}^2\to \mathbf{R}$ be defined by the equation
$$F(x,y)=f(x,y,g(x,y))$$
( a ) Find $DF$ in terms of the partials of $f$ and $g$.
( b ) If $F(x,y)=0$ for all $(x,y)$, find $D_{1}g$ and $D_{2}g$ in terms of the partials of $f$.
Solution:
( a ) We have $DF(x,y)=\left[\begin{array}{cc}{D}_{1}F(x,y)& D_{2}F(x,y)\end{array}\right]$, and
$$\begin{array}{rl}{D}_{1}F(x,y)& =\underset{t\to 0}{\lim }\frac{f(x+t,y,g(x+t,y))-f(x,y,g(x,y))}{t}\\ & =\underset{t\to 0}{\lim }\frac{f(x+t,y,g(x+t,y))-f(x,y,g(x+t,y))}{t}+\underset{t\to 0}{\lim }\frac{f(x,y,g(x+t,y))-f(x,y,g(x,y))}{t}\\ & =D_{1}f(x,y,g(x,y))+D_{3}f(x,y,g(x,y))\cdot D_{1}g(x,y)\end{array}$$
$$D_{2}F(x,y)=D_{2}f(x,y,g(x,y))+D_{3}f(x,y,g(x,y))\cdot D_{2}g(x,y)$$
( b ) In this case $D_{1}F(x,y)=D_{2}F(x,y)=0$, thus
$$D_{1}g(x,y)=[D_{3}f(x,y,g(x,y)){]}^{-1}\cdot (-D_{1}f(x,y,g(x,y)))$$
$$D_{2}g(x,y)=[D_{3}f(x,y,g(x,y)){]}^{-1}\cdot (-D_{2}f(x,y,g(x,y)))$$

Exercise 4. Let $g:{\mathbf{R}}^2\to {\mathbf{R}}^2$ be defined by the equation $g(x,y)=(x,y+x^2)$. Let $f:{\mathbf{R}}^2\to \mathbf{R}$ be the function defined in Example 3 of Section 5. Let $h=f\circ g$. Show that the directional derivatives of $f$ and $g$ exist everywhere, but there is a $\mathbf{u}\ne 0$ for which $h^{\prime }(0;\mathbf{u})$ does not exist.
Solution: The directional derivatives of f exists everywhere is clear from the text. Since $Dg(x,y)$ exists everywhere and $Dg(x,y)=\left[\begin{array}{cc}1& 2x\\ 0& 1\end{array}\right]$, the directional derivatives of g exists everywhere since
$$g^{\prime }(\mathbf{a};\mathbf{u})=Dg(\mathbf{a})\cdot \mathbf{u}$$
To show $\exists \mathbf{u}\ne 0$, s.t. $h^{\prime }(0;\mathbf{u})$ does not exist, first notice $h(0)=f(g(0))=f(0,0)=0$, let $\mathbf{u}=(1,0{)}^T$, then when $t\to 0,t\ne 0$, we have
$$\frac{h(0+t\mathbf{u})-h(0)}{t}=\frac{h(t,0)}{t}=\frac{f(t,t^2)}{t}=\frac{1}{t}\frac{t^2{t}^2}{(t^4+t^4)}=\frac{1}{2t}$$
thus the limit
$$h^{\prime }(0;\mathbf{u})=\underset{t\to 0}{\lim }\frac{h(0+t\mathbf{u})-h(0)}{t}=\underset{t\to 0}{\lim }\frac{1}{2t}$$
does not exist.