\magnification 1800
\parskip 15pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\def\N{{\bf N}}
\def\a{{\bf a}}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\ep{\epsilon}
\def\v{\vskip 10pt}
\def\u{\vskip -8pt}
\def\hb{\hfil \break}

 Let $f: \R^n \rightarrow \R$.  

Define $g_{j}: \R \rightarrow 
\R$ by
$g_{i}(t) = f_i(x_1, ..., x_{j-1}, t, x_{j+1}, ..., x_n)$.  

If $g$ is 
differentiable at $a_j$, then the partial derivative of $f$ with respect to $x_j$ is defined by 

${\partial f_i \over \partial x_j}({\bf x}) = g'(x_j) =$
$lim_{h \rightarrow 0} {g(x_j + h) - g(x_j) \over h}$
 
$= lim_{h \rightarrow 0} {f_i(x_1, ..., x_{j-1}, x_j + h, x_{j+1}, ..., x_n) - f_i(x_1, ..., 
x_{j-1}, x_j, x_{j+1}, ..., x_n) \over h}$

~~~~~~~~~$ = lim_{h \rightarrow 
0} { f_i({\bf x} + h {\bf e_j}) - f_i({\bf x}) \over h}$

\v\v\v

Ex:  ${\partial (x^2y) \over \partial x} = $ \hfil ${\partial (x^2y) \over \partial x} = $

\v

The gradient of $f$ is denoted by 
\v

\centerline{$\nabla f({\bf a}) = \left({\partial f_1 \over \partial x_1}({\bf a}), 
...,
 {\partial f_1 \over \partial x_n}({\bf a})\right)$}

\v

Ex:  $\nabla x^2y =$
\v
\v
%%\vskip 5pt
%%\hrule
%%\vskip -5pt

\eject

When is $f$ differentiable (not just partially differentiable)?

Ex:  $f: \R^2 \rightarrow \R$, $f(x, y) = \cases{ 0 & (x, y) = (0, 0) \cr {xy \over x^2 + y^2} & otherwise}$


${\partial f \over \partial x}(0, 0) = lim_{h \rightarrow 0}{f(0 + h, 0) - f(0, 0) \over h} = lim_{h \rightarrow 0}{0 - 0 \over h} = 0$ 

${\partial f \over \partial y}(0, 0) = lim_{h \rightarrow 0}{f(0, 0 + h) - f(0, 0) \over h} = lim_{h \rightarrow 0}{0 - 0 \over h} = 0$ 


BUT $f$ is not continuous at (0, 0)!!!!!!!!!


RECALL:  $f$ differentiable implies $f$ continuous
\vfill
\vfill
\vskip 50pt
\hrule
\vskip -5pt

Ex:  $g: \R^2 \rightarrow \R$, $g(x, y) = { x + 2y }$


${\partial g \over \partial x}(0, 0) = lim_{h \rightarrow 0}{f(0 + h, 0) - f(0, 0) \over h} = lim_{h \rightarrow 0}{h  - 0 \over h} = 1$ 

${\partial g \over \partial y}(0, 0) = lim_{h \rightarrow 0}{f(0, 0 + h) - f(0, 0) \over h} = lim_{h \rightarrow 0}{2h - 0 \over h} = 2$ 


We will see later that $g$ is differentiable.
\eject

Suppose $f: \R^1 \rightarrow \R^1$ is differentiable.  Then
{$$lim_{h \rightarrow 0}{f(a+h) - f(a) \over h} = f'(a)$$} 
{$$ lim_{h \rightarrow 0}{f(a+h) - f(a) \over h} - f'(a) = 0$$}
$$lim_{h \rightarrow 0}{f(a+h)  - f(a) - f'(a)h\over h} = 0$$
or equivalently,
$$lim_{x \rightarrow a}{f(x) - f(a) \over x-a} = f'(a)$$ 
$$ lim_{x \rightarrow a}{f(x) - f(a) \over x-a} - f'(a) = 0$$
$$lim_{x \rightarrow a}{f(x)  - f(a) - f'(a)(x-a)\over (x-a)} = 0$$
$$lim_{x \rightarrow a}{f(x)  - [f(a) + f'(a)(x-a)]\over (x-a)} = 0$$

Thus when $x$ is close to $a$, then $f(x)$ is close to $f(a) + f'(a)[x-a]$ 

{Thus $y = f(a) + f'(a)[x-a]$ is the best linear approximation of $f$ near $x = a$.} 

\vfill
\vfill
\vfill
\eject
Defn:  Let $V$ and $W$ be vector spaces.  A {\bf linear transformation} 
from $V$ to
$W$ is a function $T: V \rightarrow W$ that satisfies the following 
two 
conditions.
For each {\bf u} and {\bf v} in $V$ and scalar $a$,

i.)  $T(a{\bf u}) = aT({\bf u})$

ii.)  $T({\bf u + v}) = T({\bf u}) + T({\bf v})$

%%In this case, we say, $T \in {\cal L}(R^n;R^m)$.
\vskip 5pt
\hrule


Thm:  Let $T: V \rightarrow W$ be a linear transformation.  Then
$T({\bf 0}) = {\bf 0}$

Pf:  $T({\bf 0}) = T({\bf 0} +  {\bf 0}) = T({\bf 0}) +  T({\bf 0})$

\vskip 5pt
\hrule

Thm:  Let $A$ be an $m \times n$ matrix.  Then the function
$$\matrix{ T:\R^n \rightarrow \R^m \cr 
T({\bf x}) = A{\bf x}}$$ is a linear transformation.

Thm:  If $T: \R^n \rightarrow \R^m$ is a linear transformation,
then $T({\bf x}) = A{\bf x}$ where $$A = [T({\bf e_1}) ... T({\bf e_n})]$$

Ex:  If $T: \R^2 \rightarrow \R$ and $T(1, 0) = 3$, $T(0, 1) = 4$, then 
$$T(x, y) = xT(1, 0) + yT(0, 1) = (3 ~~4) \left( \matrix{x \cr y} \right) = 3x + 4y$$

\vfill

\eject

Defn:  Suppose $f: \R^1 \rightarrow \R^1$ is differentiable at a. Then
 $$lim_{h \rightarrow 0}{f(a+h)  - f(a) - f'(a)h\over h} = 0$$
$$lim_{x \rightarrow a}{f(x)  - f(a) - f'(a)(x-a)\over (x-a)} = 0$$
$$lim_{x \rightarrow a}{f(x)  - [f(a) + f'(a)(x-a)]\over (x-a)} = 0$$
Thus when $x$ is close to $a$, then $f(x)$ is close to $f(a) + f'(a)[x-a]$ 

Thus $y = f(a) + f'(a)[x-a] $ is the best linear approximation of $f$ near $x = a$. 

Hence the tangent line to $f$ at $a$ is $y = f(a) + f'(a)[x-a]$ 

The tangent line can be written as a constant plus a linear function.

\vskip 5pt
\hrule
\vskip -5pt



Defn:  Suppose $A \subset \R^n$, $f: A \rightarrow \R^m$.  
\vskip -5pt
$f$ is said to be {\bf differentiable at a point a} if there exists 
an open ball $V$ such that $a \in V \subset A$ and a linear function $T$ 
%%$\in {\cal L}(R^n; R^m)$ 
such that
$$lim_{{\bf h} \rightarrow {\bf 0}}{||f({\bf a}+{\bf h})  - f({\bf a}) - 
T({\bf h})||\over ||{\bf h}||} = 0$$
$$lim_{{\bf x} \rightarrow {\bf a}}{||f({\bf x}) - f({\bf a}) - 
T({\bf x} - {\bf a})||\over ||{\bf x} - {\bf a}||} = 0$$

Suppose $f:  \R^n \rightarrow \R$.  \hfil \break
Then $T = ( b_1 ~~b_2 ~... ~b_n)$ and $T{\bf x} = 
(b_1 ~~b_2 ~... ~ b_n) \left( \matrix{x_1 \cr x_2 \cr .\cr .\cr . \cr x_n} \right) = \Sigma b_ix_i$ 

Also, $T({\bf x} - \a) = 
(b_1 ~~b_2 ...~ b_n) \left( \matrix{x_1 - a_1 \cr x_2 - a_2 \cr . \cr . \cr . \cr x_n - a_n} \right) = \Sigma b_i(x_i - a_i)$ 


$$lim_{{\bf x} \rightarrow {\bf a}}{||f({\bf x}) - f({\bf a}) - 
T({\bf x} - {\bf a})||\over ||{\bf x} - {\bf a}||} = 0$$

$$lim_{{\bf x} \rightarrow {\bf a}}{||f({\bf x}) - f({\bf a}) - 
 \Sigma b_i(x_i - a_i)||\over ||{\bf x} - {\bf a}||} = 0$$


$$lim_{{\bf x} \rightarrow {\bf a}}{f({\bf x}) - f({\bf a}) - 
 \Sigma b_i(x_i - a_i) \over ||{\bf x} - {\bf a}||} = 0$$

$y =  f({\bf a}) + 
 \Sigma b_i(x_i - a_i)$ approximates $y = f({\bf x})$


\end
 \eject




















Defn:  Suppose $f: \R^1 \rightarrow \R^1$ is differentiable at a. Then

 $$lim_{h \rightarrow 0}{f(a+h)  - f(a) - f'(a)h\over h} = 0$$

$$lim_{x \rightarrow a}{f(x)  - f(a) - f'(a)(x-a)\over (x-a)} = 0$$

Defn:  Suppose $A \subset \R^n$, $f: A \rightarrow \R^m$.  
\vskip -5pt
$f$ is said to be {\bf differentiable at a point a} if there exists 
an open ball $V$ such that $a \in V \subset A$ and a linear function $T$ 
%%$\in {\cal L}(R^n; R^m)$ 
such that
$$lim_{{\bf h} \rightarrow {\bf 0}}{||f({\bf a}+{\bf h})  - f({\bf a}) - 
T({\bf h})||\over ||{\bf h}||} = 0$$

$$lim_{{\bf x} \rightarrow {\bf a}}{||f({\bf x}) - f({\bf a}) - 
T({\bf x} - {\bf a})||\over ||{\bf x} - {\bf a}||} = 0$$

Suppose $f:  \R^n \rightarrow \R$.  \hfil \break
Then $T = ( b_1, ..., b_n)$ and $T{\bf x} = 
(b_1, ..., b_n) \left( \matrix{x_1 \cr x_1 \cr ... \cr x_n} \right) = \Sigma b_ix_i$ 

$$lim_{{\bf x} \rightarrow {\bf a}}{||f({\bf x}) - f({\bf a}) - 
 \Sigma b_i(x_i - a_i)||\over ||{\bf x} - {\bf a}||} = 0$$

\eject

$f: \R \rightarrow \R$:
\vskip -5pt

$$lim_{x \rightarrow a}{f(x)  - f(a) - f'(a)(x-a)\over (x-a)} = 0$$


$y = f'(a)[x-a] + f(a)$ is the linear approximation of $f$ near $a$.

${df \over dx}(a) = f'(a)$

\vskip 5pt
\hrule

$f: \R^n \rightarrow \R$:
\vskip -5pt

$$lim_{{\bf x} \rightarrow {\bf a}}{f({\bf x}) - f({\bf a}) - 
 \Sigma b_i(x_i - a_i) \over ||{\bf x} - {\bf a}||} = 0$$

$y =  f({\bf a}) + 
 \Sigma b_i(x_i - a_i)$ approximates $y = f({\bf x})$

$f({\bf x}) =  f({\bf a}) +
 \Sigma b_i(x_i - a_i)  + ||{\bf x} - {\bf a}|| r({\bf x}, {\bf a})$ 

\rightline{where $lim_{{\bf x} 
\rightarrow {\bf a}} r({\bf x}, {\bf a}) = 0$}

$(df)_a =  \Sigma b_i(x_i - a_i)  $
\vskip 5pt
\hrule

Mean Value Theorem:  Suppose \hfil \break
1.) $f$ continuous on $[a, b]$\hfil \break
2.) $f$ differentiable on $(a, b)$
\vskip -5pt
Then there exists $c \in (a, b)$ such that $f'(c) = {f(b) - f(a) \over b - a}$



\vskip 5pt
\hrule


http://www.geometrygames.org/ 
\vskip -10pt
  http://www.geometrygames.org/SoS/

\eject

$f: \R^n \rightarrow \R$ is differentiable at $a$ if

\vskip -5pt

$$lim_{{\bf x} \rightarrow {\bf a}}{f({\bf x}) - f({\bf a}) - 
 T({\bf x}-  {\bf a}) \over ||{\bf x} - {\bf a}||} = 0$$


$$lim_{{\bf x} \rightarrow {\bf a}} {f({\bf x}) - f({\bf a}) - 
 \Sigma b_i(x_i - a_i) \over ||{\bf x} - {\bf a}||} = 0$$

$y =  f({\bf a}) + 
 \Sigma b_i(x_i - a_i)$ approximates $y = f({\bf x})$

$f({\bf x}) =  f({\bf a}) +
 T({\bf x}-  {\bf a})   + ||{\bf x} - {\bf a}|| r({\bf x}, {\bf a})$ 

\rightline{where $lim_{{\bf x} 
\rightarrow {\bf a}} r({\bf x}, {\bf a}) = 0$}


Thm 1.1:  If $f$ is differentiable at $a$, then 

1.)  $f$ is continuous at $a$.

2.)  All partial derivatives exist at $a$.

3.)  $b_i = ( {\partial f \over \partial x_i})_a$

Proof:  1.) $lim_{{\bf x} \rightarrow {\bf a}}f({\bf x}) =  lim_{{\bf x} \rightarrow {\bf a}}f({\bf a}) +
 T({\bf x}-  {\bf a})   + ||{\bf x} - {\bf a}|| r({\bf x}, {\bf a}) =$ 

2,3.)  ${\partial f \over \partial x_j}({\bf a}) = lim_{h \rightarrow 
0} { f({\bf a} + h {\bf e_j}) - f({\bf a}) \over h}$


Thm 1.3: If ${\partial f \over \partial x_j}$ exist for all $j$ in a nbhd of $a$ and if they are continuous at $a$, then $f$ is differentiable at $a$.

Proof:  HW

Defn: Let  $V$ be a nonempty open subset of $R^n$, $f: V \rightarrow R^m$, 
$p \in {\N}$.

i.)  $f$ is $C^p$ on $V$ is each partial derivative of order $k \leq p$ 
exists and is continuous on $V$.

ii.)  $f$ is $C^\infty$ on $V$ if  $f$ is $C^p$ on $V$ for all $p \in 
{\N}$
\vfill
\vskip 5pt
\hrule
\vfill

Chain rule 1:  Suppose $f: (a, b) \rightarrow \R^n$, $g: \R^n \rightarrow \R$, then 

${d \over dt} (g \circ f)_{t_0} = D(G)_{f(t_0)} D(f)_{t_0} = (b_1, ..., b_n)
 \left(\matrix{
f_1'(t_0) \cr f_2'(t_0) \cr ...\cr f_n'(t_0)}\right) $
\vskip 15pt
\rightline{$=   \S ({\partial g \over \partial x_i})_{f({t_0})}
({df_i \over dt})_{t_0}$}

Ex:  $f(t) = (t^2, sin(t))$, $D(f) =  \left(\matrix{
2t \cr cos(t) }\right)
 
$g(x, y) =  x + y^3$, $D(g) = (1, 3y^2)$


$(g \circ f)({t}) = g(t^2, sin(t)) = t^2 + sin^3(t)$   

$(g \circ f)'({t}) = 2t_0 + 3sin^2(t_0)cos(t_0)$

$ D(g)_{f(t_0)} D(f)_{t_0} = (1, 3sin^2(t_0))\left(\matrix{
2t_0 \cr cos(t_0) }\right) 
$,  

Defn:  $U$ is starlike with respect to {\bf a} if ${\bf x} \in U$ implies $\overline{\bf ax} \subset U$

Thm 1.5 (Mean Value Theorem)  Let $g$ by a 
Suppose $U$ is starlike with respect to {\bf a} .  Then given ${\bf x} \in U$, there exists $c \in \R$, $0 < t < 1$ such that 

$g({\bf x}) - g({\bf a}) = \S ({\partial g \over \partial x_i})_{\bf p} (x_i - a_i)$ 

where ${\bf p} = {\bf a} + t({\bf x} - {\bf a})




















Defn:  The {\bf Jacobian matrix of $f$ at a} is

$$\left[\matrix{{\partial f_i \over \partial x_j}({\bf a})}\right]_{m 
\times n}
=
\left[\matrix{
{\partial f_1 \over \partial x_1}({\bf a}) & ... 
& {\partial f_1 \over \partial x_n}({\bf a}) \cr
. & .~~~~ & . \cr
. & ~~.~~ & . \cr
. & ~~~~. & . \cr
{\partial f_m \over \partial x_1}({\bf a}) & ... 
& {\partial f_m \over \partial x_n}({\bf a}) \cr
}\right]$$



Thm 11.12:   Let  $V$ be a nonempty open subset of $R^n$, $f: V 
\rightarrow R^m$.  If $f$ is differentiable at some point ${\bf a} \in V$, 
then all first order partial derivatives of $f$ exist at {\bf a} and the 
matrix which represents the total derivative of $f$ at ${\bf a}$ is the 
Jacobian matrix of $f$ at ${\bf a}$.

If $m = 1$, $Df({\bf a}) = \left[
{\partial f_1 \over \partial x_1}
({\bf a})  
...
 {\partial f_1 \over \partial x_n}({\bf a})\right]$

In this case, the gradient of $f$ is denoted by 
$$\nabla f({\bf a}) = \left({\partial f_1 \over \partial x_1}({\bf a}), 
...,
 {\partial f_1 \over \partial x_n}({\bf a})\right)$$




Thm H:  Let $V$ be an open subset of $R^n$, ${\bf a} \in V$, $f: V
\rightarrow R^m$. $f = (f_1, ..., f_m)$ is differentiable at {\bf a} if 
and only if for all $i$,
 $$lim_{{\bf h} \rightarrow 0}
{f_i({\bf a} + {\bf h}) - f_i({\bf 
a}) - \nabla f_i({\bf
a}) \cdot {\bf h} \over ||{\bf h}||} = 0$$

Ex:  The Jacobian of $f(x, y) = \cases{ x + y & if $xy = 0$ \cr
1 & otherwise}$ exists, but $f$ is not differentiable at (0, 0).
\vskip 8pt

Thm:  Let $V$ be an open subset of $R^n$, ${\bf a} \in V$, $f: V
\rightarrow R^m$.  If all first order partial derivatives of $f$ exist in 
$V$ and are continuous at {\bf a}, then $f$ is differentiable at {\bf a}.

\vfill
\vskip 5pt
\hrule
\vfill

Lemma 11.21:   Let $V$ be an open subset of $R^n$, ${\bf a} \in V$, $f: V
\rightarrow R^m$. Then $f$ is differentiable at {\bf a} if and only if 
there is a linear function $T \in {\cal L}(R^n; R^m)$ and a function 
$\epsilon: R^n \rightarrow R^m$ such that $lim_{{\bf h} \rightarrow 
0} \epsilon({\bf h}) = {\bf 0}$ and $$f({\bf a} + {\bf h}) - f({\bf 
a}) = T({\bf h}) + ||{\bf h}|| \epsilon({\bf h})$$ for sufficiently small 
$h$ (i.e., there exists an $r > 0$ such that the above holds for $||{\bf 
h}|| < r$).

Lemma 11.22: $D(f \cdot g)({\bf a}) = g({\bf a})Df({\bf a}) + f({\bf a})
Dg({\bf a})$, etc.
\end



\end





\end