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\newtheorem{thm}{Theorem}[section]
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\title{Differentiable Functions}
\author{Ericka L. Sunnerville}
\date{February 2004}

\begin{document}

\maketitle

\begin{abstract}
In this paper I will briefly discuss one of the two fundamental operations of calculus-differentiation.
\end{abstract}

\tableofcontents

\section{Introduction}
\label{sec:introduction}

In this paper I will give a definition of the derivative of a function, state some basic rules for differentiation, and give some examples of functions that are and are not differentiable

\section{Definition}
The \textbf{derivative} of $f$ at $x$ is give by

\begin{displaymath}
f'(x) = \lim_{\Delta x \mapsto 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}
\end{displaymath}

provided the limit exists.

\textbf{Note}: The process of finding the derivative of a function is called \textit{differentiation}.  A function is \textit{differentiable} at $x$ if its derivative exists at $s$ and \textit{differentiable} on an open interval $(a,b)$ if it is differentiable at every point in the interval.

Other notations for the derivative of $f$ include: $D_y[f(x)]$, $\frac{dy}{dx}$, or 
$y'$. The last two are used when the rule for $f$ is written in the form $y = f(x)$.

\section{Calculating the Derivative}
The calculation of the derivative of $f$ is facilitated using the following four-step process.
\begin{enumerate}
	\item Compute  $f(x + \Delta x)$.
	\item Form the $difference$  $f(x + \Delta x) - f(x)$.
	\item Form the $quotient$  $\frac{f(x + \Delta x) - f(x)}{\Delta x}$.
	\item Compute  $f'(x) = \lim_{\Delta x \mapsto 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$.
\end{enumerate}



\begin{ex}
Let $f(x) = x^2$. Compute $f'(x)$.

\begin{eqnarray*}
f(x + \Delta x) 
   &=& (x + \Delta x)^2 = x^2 + 2x\Delta x + {\Delta x}^2 \\
f(x + \Delta x) - f(x) 
   &=& x^2 + 2x\Delta x + {\Delta x}^2 - x^2 = 2x\Delta x + {\Delta x}^2 
    = \Delta x (2x + \Delta x) \\
\frac{f(x + \Delta x) - f(x)}{\Delta x} 
   &=& \frac{\Delta x (2x + \Delta x)}{\Delta x} = 2x + \Delta x \\
f'(x) 
   &=& \lim_{\Delta x \mapsto 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \mapsto 0}(2x + \Delta x) = 2x
\end{eqnarray*}

\end{ex}

\section{Differentiability and Continuity}
In practical applications, you encounter functions that fail to be \textit{differentible}-that is, do not have a derivative at certain values in the domain of the function $f$.  It can be shown that a continuous function $f$ fails to be differentiable at a point when the graph of $f$ makes and abrupt change of direction at that point.  The following alternative limit form of the derivative is useful un investigating the relationship between differentiability and continuity.  The derivative of $f$ at $c$ is

\begin{displaymath}
f'(c) = \lim_{x \mapsto c} \frac{f(x) - f(c)}{x - c}
\end{displaymath}
provided this limit exists.  Note that the limit in this alternative form requires that the one-sided limits

\begin{displaymath}
\lim_{x \mapsto c^{+}} \frac{f(x) - f(c)}{x-c}  
\hbox{   and   }
\lim_{x \mapsto c^{-}} \frac{(f(x)-f(c)}{x-c}
\end{displaymath}

exists and are equal.  These one-sided limits are called the \textbf{derivatives from the left and from the right}, respectively.

\begin{ex}
The function
\begin{displaymath}
f(x) = |x - 2|
\end{displaymath}
is continuous at x = 2.  However, the one-sided limits

\begin{displaymath}
lim \frac{f(x) - f(2)}{x - 2} = lim \frac{|x - 2| - 0}{x - 2} = -1  \quad and \quad
lim \frac{f(x) - f(2)}{x - 2} = lim \frac{|x - 2| - 0}{x - 2} = 1
\end{displaymath}
are not equal.  Therefore, $f$ is not differentiable at $x = 2$.

\end{ex}

\begin{thm}{Differentiability Implies Coninuity}

If $f$ is differentiable at $x = c$, then $f$ is continuous at $x = c$.
\end{thm}
\proof
You can prove that $f$ is continuous at $x = c$ by showing that $f(x)$ approaches $f(c)$ as $x\rightarrow c$.  To do this, use the differentiability of $f$ at $x\rightarrow
c$ and consider the following limit:

\begin{eqnarray*}
lim[f(x) - f(c)] &=& lim [(x - c)(\frac{f(x) - f(c)}{x - c})] \\
								 &=& [lim (x - c)][lim \frac{f(x) - f(c)}{x - c}] \\
								 &=& (0)[f'(c)] \\
                 &=& 0
\end{eqnarray*}

Because the difference $f(x) - f(c)$ approaches zero as $x\rightarrow c$, you can conclude that $lim f(x) = f(c)$.  Therefore, $f$ is continuous at $x = c$.
\proofend

\textbf{Note:} If a function is differentiable at $x = c$, then it is continuous at $x = c$.  Hence, differentiability implies continuity.  It is possible for a function to be continuous at $x = c$ and \textit{not} be differentiable at $x = c$.  Hence, continuity does not imply differentiability.

\section{A few Theorems for Differentiable Functions}

There are many theorems related to differentiable functions. We already saw, for example, that differentiablity implies continuity. Other important theorems are \textit{Rolle's Theorem} and its generalized form.

\begin{thm}{Rolle's Theorem}
Let $f$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$.  If $f(a) = f(b)$ then there is at least one number $c$ in $(a,b)$ such that $f'(c)=0$.
\end{thm}

\begin{ex}
\label{ex:rolle1}
Suppose $f(x) = x^4 - 2x^2$ and we consider the interval $[-2,2]$.  Does Rolle's Theorem apply, and if so find the points guaranteed to exist by the theorem.
\end{ex}

\begin{figure}
			\centering
			\includegraphics[width=0.50\textwidth]{graph.eps}
			\caption{Graph of $f(x) = x^4 - 2x^2$}
			\label{fig:cubic}
\end{figure}

To begin, note that $f$ is continuous on the interval $[-2,2]$ and differentiable on the interval $(-2,2)$.  Moreover, because $f(-2) = 8 = f(2)$ you can conclude that there exists at least one $c$ in $(-2,2)$ such that $f'(c) = 0$.  Setting the derivative equal to $0$ produces

\begin{displaymath}
f'(x) = 4x^3 - 4x = 0
\end{displaymath}
\begin{displaymath}
4x(x^2-1) = 0
\end{displaymath}
\begin{displaymath}
x=0,1,-1
\end{displaymath}


\begin{thm}{Generalized Rolle's Theorem}
\begin{itemize}
\item $f$ once differentiable, $f(a_1) = f(a_2)$ $\Rightarrow$ $f'(c_1) = 0$ for some $c_1$
\item $f$ twice differentiable, $f(a_1) = f(a_2) = f(a_3)$ 
 \begin{itemize}
 \item $\Rightarrow$ $f'(c_1) = 0$ and $f'(c_2) = 0$
 \item $\Rightarrow$ $f''(d_1) = 0$ for some $d_1$
 \end{itemize}
\item $f$ thrice differentiable, $f(a_1) = f(a_2) = f(a_3) = f(a_4)$ 
 \begin{itemize}
 \item $\Rightarrow$ $f'(c_1) = 0$ , $f'(c_2) = 0$, and $f'(c_3) = 0$
 \item $\Rightarrow$ $f''(d_1) = 0$ and $f''(d_2) = 0$
 \item $\Rightarrow$ $f'''(e_1)=0$ for some $e_1$
 \end{itemize}
\end{itemize}
\end{thm}

\begin{ex}
let $f(x) = x^4 - 2x^2$ on $[-1.3,1.3]$ then by the Generalized Rolle's Theorem, $f(1.3) = -0.5239 = f(-1.3)$.
\begin{displaymath}
f'(x) = 4x^3 - 4x = 0
\end{displaymath}
\begin{displaymath}
4x(x^2-1) = 0
\end{displaymath}
\begin{displaymath}
x=0,1,-1
\end{displaymath}
Thus there are three points where $f'(x) = 0$.     								 

\begin{displaymath}
f''(x) = 12x^2 - 4 = 0
\end{displaymath}
\begin{displaymath}
12x^2 = 4
\end{displaymath}
\begin{displaymath}
x= \frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}
\end{displaymath}
Thus there are two points where $f''(x) = 0$

\begin{displaymath}
f'''(x) = 24x = 0
\end{displaymath}
\begin{displaymath}
x = 0
\end{displaymath}
Thus there is one point where $f'''(x) = 0$

\end{ex}



Now we saw that functions that are differentiable more than once could be useful, we define new classes of functions as follows.

\begin{dfn}{The Space of $C^n$ Functions}
If $f(x)$ is $n$ times differentiable on $[a,b]$ and if $f^{n}(x)$ is itself continuous on $[a,b]$, we shall write $f(x) \in C^n[a,b]$.
\end{dfn}

\textbf{Note:}$C^n[a,b]$ is a linear space of functions.


\begin{ex}Here is a function that is continuous but not differentiable (i.e. $C^0$):

Let $f(x) = |x|$ and we consider the interval $[-1,1]$.  The function is certainly continuous, but not differentiable at $x = 0$.\\  
\end{ex}


\begin{ex}Here is a function that is differentiable but $f'$ is not continuous (i.e. again $C^0$):
\begin{displaymath}
f(x) = \left\{ 
\begin{array}{cc}
x^2 sin( {1 \over x} ) &  x \ne 0 \\
0                    &  x = 0 \\
\end{array}
\right.
\end{displaymath}	
\end{ex}

\begin{ex} The function $f(x) = x^{4 \over 3}$ is a function that is in $C^1$ but not $C^2$.
\begin{displaymath}
f(x) = x^{4 \over 3}, x \in \Re
\end{displaymath}

\begin{displaymath}
f'(x) = \frac{4}{3} x^{1 \over 3} \Rightarrow f \in C'
\end{displaymath}

\begin{displaymath}
f''(x) = \frac{4}{3} \cdot \frac{1}{3}x^{\frac{-2}{3}}
\end{displaymath}

\begin{displaymath}
f''(x) = \frac{4}{9} \cdot \frac{1}{x^{2 \over 3}}.  
\end{displaymath}
The function is not defined at $x = 0$, and thus not in $C^2$.

\end{ex}

\section{Taylor's Theorem}

Now our basic question is this: if a function $f$ is $n$-times differentiable, i.e.
$f \in C^n$, then what does the function have to do with its derivatives, if anything?
Or in other words, can I use the derivatives to approximate the function.  The answer is yes, and Taylor's theorem helps us do so.

\begin{thm}{Taylor's Thereom}
Let $f(x) \in C^{n + 1}[a,b]$ and let $x_0 \in [a,b]$, then 
\begin{displaymath}
f(x)\approx f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \ldots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + \frac{1}{n!} \int_{t=x_0}^{t=x} f^{(n+1)}(t)(x-t)^n dt.
\end{displaymath}
\end{thm}

\begin{thm}{Taylor's Theorem with Exact Remainder}
Let $f(x) \in C^{n}[a,b]$ and let $f^{(n+1)}(x)$ exists in $(a,b)$. Then there exists a $\zeta$ with $ a < \zeta < b$ such that
\begin{displaymath}
f(b) = f(a) + f'(a)(b - a) + \frac{f''(a)}{2!}(b - a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(b - a)^n + \frac{f^{(n + 1)}\zeta}{(n + 1)!}(b - a)^{n+1}.
\end{displaymath}

\end{thm}

\begin{ex}
Let $f(x) = e^x \in C^5$ for $x = 2$ , on $[-1,1]$. $f(x)$ is differentiable infinitely many times and since $f^{(n)}(x) = e^x$ for all $n$, $f^{(n)}{x_0} = 1$. The Taylor expansion is given by

\begin{displaymath}
e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 + \frac{e^{\zeta_5}}{720}x^6 = 7.26667 + some\; arbitrary\; number
\end{displaymath}
From Taylor's Theorem, it is now easy to obtain a polynomial approximation by deleting the remainder term from the formula. Now we have

\begin{displaymath}
e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 = 7.26667  
\end{displaymath}

$e^2$ = 8.26667.  Taylor's theorem gives a close approximation to the value of a function which we can not easily compute.

\end{ex}




\end{document}