<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">%% LyX 1.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[11pt,oneside]{amsart}
\usepackage[T1]{fontenc}
\usepackage{geometry}
\geometry{verbose,letterpaper,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
\setlength\parskip{\medskipamount}
\setlength\parindent{0pt}

\makeatletter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
\providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
 \theoremstyle{plain}    
 \newtheorem{thm}{Theorem}[section]
 \numberwithin{equation}{section} %% Comment out for sequentially-numbered
 \numberwithin{figure}{section} %% Comment out for sequentially-numbered
 \theoremstyle{remark}
 \newtheorem*{rem*}{Remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\setcounter{section}{2}
\setcounter{subsection}{1}

\makeatother

\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Differentiation Rules}

\begin{thm}
\textbf{\textup{{[}The Chain Rule{]}}}  \( \left( f(g(x))\right) '=f'(g(x))\cdot g'(x). \)
\end{thm}
\begin{proof}
Again, we appeal to the definition of the derivative: \begin{eqnarray*}
\left( f(g(x))\right) ' &amp; := &amp; \lim _{h\rightarrow 0}\frac{f(g(x+h))-f(g(x))}{h}\\
 &amp; = &amp; \lim _{h\rightarrow 0}\frac{(f(g(x+h))-f(g(x)))(g(x+h)-g(x))}{h(g(x+h)-g(x))}.
\end{eqnarray*}
Now, we do some clever substitutions; som e may say sleight-of-hand substitutions.
The idea here is to get you to look at the same term two different ways. Substitute
\( u=g(x) \), and \( u+\Delta u=g(x+h) \) (so that \( \Delta u=g(x+h)-g(x) \)),
and: \begin{eqnarray*}
\left( f(g(x))\right) ' &amp; = &amp; \lim _{h\rightarrow 0}\frac{(f(g(x+h))-f(g(x)))(g(x+h)-g(x))}{h(g(x+h)-g(x))}\\
 &amp; = &amp; \lim _{h\rightarrow 0}\frac{(f(u+\Delta u))-f(u))(g(x+h)-g(x))}{h(\Delta u)}\\
 &amp; = &amp; \left( \lim _{\Delta u\rightarrow 0}\frac{f(u+\Delta u))-f(u)}{\Delta u}\right) \left( \lim _{h\rightarrow 0}\frac{g(x+h)-g(x))}{h}\right) \\
 &amp; = &amp; f'(u)\cdot g'(x)=f'(g(x))\cdot g'(x).
\end{eqnarray*}
Note how you are looking at \( \Delta u \) in the first difference quotient
as the change in the variable, but in the second difference quotient it changes
roles to the difference of the values of \( g \). Now, also, I changed the
first limit to as \( \Delta u\rightarrow 0 \). This is justifiable, since as
\( h\rightarrow 0 \), \( \Delta u\rightarrow 0 \). 
\end{proof}
\begin{rem*}
There is one thing wrong with this proof, though, and that is that it \textit{might}
be that \( g(x+h)=g(x) \) lots of times as \( h \) goes down to 0, so the
limits as written won't make much sense. That is, while certainly \( \Delta u\rightarrow 0 \)
when \( h\rightarrow 0 \), it might be that \( \Delta u \) goes to 0 well
before \( h \) does, so you can't quite say that \( h\rightarrow 0 \) when
\( \Delta u\rightarrow 0 \). However, if that happens, that is, if \( \Delta u \)
goes to zero before \( h \) does, or hits 0 many times as \( h\rightarrow 0 \),
then \( g'(x)=0 \), as will be \( \left( f(g(x))\right) ' \), so the formula
works even in this case. There is another proof which works around this difficulty,
but this is the proof that makes the most sense, despite this theoretical question
mark.\end{rem*}

\end{document}
</pre></body></html>