\magnification 1800
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\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\f{\vskip 10pt}
\def\u{\vskip -8pt}

Defn:  Let $f: \R^n \rightarrow \R^m$.  Define $g_{ij}: \R^1 \rightarrow 
\R^1$, \hfil \break 
$g_{ij}(t) = f_i(x_1, ..., x_{j-1}, t, x_{j+1}, ..., x_n)$.  If $g$ is 
differentiable at $a$, then the partial derivative of $f_i$ is defined by 

${\partial f_i \over \partial x_j}({\bf a}) = $ 
$lim_{h \rightarrow 0} {f_i(a_1, ..., a_{j-1}, a_j + h, a_{j+1}, ..., a_n) - f_i(a_1, ..., 
a_{j-1}, a_j, a_{j+1}, ..., a_n) \over h}$

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

\vskip 5pt
\hrule
\vskip -5pt

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)!!!!!!!!!

\vskip 5pt
\hrule
\vskip -5pt

Defn:  Suppose $f: \R^1 \rightarrow \R^1$.  The derivative of $f$ at the 
point $a$ is
{$f'(a) = lim_{h \rightarrow 0}{f(a+h) - f(a) \over h}$} if above limit exists.

$f'(a)$ is the derivative of $f$ at $x$ if 
 $$lim_{h \rightarrow 0}{f(a+h)  - f(a) - f'(a)h\over h} = 0$$

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


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

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

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\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

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

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

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 $a$.

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

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\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)  $
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\hrule

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



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\hrule


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

\end