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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle

\subsection{Differentiation Rules}

There are a number of general ``rules'' which are applied to find the derivative
of an elementary function. Here is the first of these rules: 

\begin{thm}
The derivative of a constant is 0, \( (c)'=0. \)
\end{thm}
Deep result. A bit more seriously, 

\begin{thm}
\textbf{\emph{{[}The Power Rule{]}.}} For each positive integer \( n \), \( (x^{n})'=nx^{n-1}. \) 
\end{thm}
\begin{proof}
~
\end{proof}
Even though we haven't yet proved the fact, this rule works for all powers of
\( x \), not just integer powers. For example, we saw before that \[
(\sqrt{x})'=\frac{1}{2\sqrt{x}},\]
which  translates to \[
(x^{1/2})'=(1/2)x^{\left( \frac{1}{2}-1\right) },\]
 just fitting the same rule.  

\begin{thm}
\textbf{\textup{{[}The Power Rule{]}.}} For any fixed exponent \( a \),  \( (x^{a})'=ax^{a-1} \)
on any interval on which both \( x^{a} \) and \( x^{a-1} \) are both defined.
\end{thm}
What is that stuff about \( x^{a} \) and \( x^{a-1} \) being defined? Well,
if, say, \( a=1/2 \), then \( x^{a} \) will only be defined for \( x\geq 0 \),
and \[
x^{a-1}=x^{-1/2}\]
 will only be defined for \( x>0 \). That is the sort of restriction that part
of the statement is worried about.

\begin{thm}
The derivative of a sum is is the sum of the derivatives, \( (f+g)'=f'+g'. \)
Similarly, the The derivative of a difference is is the difference of the  derivatives,
\( (f-g)'=f'-g'. \)
\end{thm}
Again, this is too trivial to prove. There is, nonetheless, a proof in the text,
but students get confused on theorems like this because it's not really clear
what you have to prove and what is ``given''.  


\subsection{The Product and Quotient Rules}

These next results are worthy of proof. And, for those who haven't memorized
these rules, they also need proof, since they aren't what you'd expect ---  unless,
as I said, someone already made you learn them.  

\begin{thm}
\textbf{\textup{{[}The Product Rule{]}}} The derivative of a product \( fg \),
is given by \[
(f\cdot g)'=f'\cdot g+f\cdot g'.\]

\end{thm}
\begin{proof}
~
\end{proof}
Of course, this theorem and the first theorem can be combined, to show that
 \( (c\cdot f)'=c\cdot f'. \) But I wouldn't call that a theorem. It is, still,
useful to remember. I think of it as just ``constants go along for the ride''. 

\begin{example}
These rules are all we need to differentiate any polynomial \[
p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}.\]
You differentiate it term by term (using the sum rule), and use the power rule
on each term (each \emph{monomial}), using this constant case of the product
rule to bring the constants along for the ride:\[
p'(x)=na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots +2a_{2}x+a_{1}.\]
As a specific example:\begin{eqnarray*}
\left( x^{4}-3x^{3}+5x^{2}-2x+1\right) ' & = & 4x^{3}-3(3x^{2})+5(2x)-2(1)+0\\
 & = & 4x^{3}-9x^{2}+10x-2.
\end{eqnarray*}

\end{example}
Here are some more examples and exercises just using these rules:

\begin{example}
Find \[
\left( \left( x^{2}+3x-2\right) \left( 3x^{4}+x^{2}-1\right) \right) '.\]

\end{example}

\subsubsection*{Solution}

\begin{example}
Find \[
\frac{d}{dx}\left( \left( x^{2}+1\right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) \right) .\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find \[
\frac{d}{dx}\left( x(x^{2}+1)(2x^{3}+3x-2)\right) .\]

\end{xca}

\subsubsection*{Answer:}

\begin{xca}
Find \[
\frac{d}{dx}\left( (x^{5}+3x^{4}-7x^{3}+x^{2}+1)(x^{5}-3x^{4}-2x^{3}-x+3)\right) .\]

\end{xca}

\subsubsection*{Answer:}

\begin{note}
Submit your answers to each group of exercises on this page separately. This
button only sends your answers to these two exercises.
\end{note}
Our next big rule is the quotient rule: 

\begin{thm}
\textbf{\textup{{[}The Quotient Rule{]}}} The derivative of a quotient \[
\frac{f}{g},\]
 is given by\[
\left( \frac{f}{g}\right) '=\frac{f'\cdot g-g'\cdot f}{g^{2}}.\]

\end{thm}
\begin{proof}
~
\end{proof}
\begin{example}
Find the derivative of \[
f(x)=\frac{x^{2}+2}{x^{2}+3x+3}\]

\end{example}

\subsubsection*{Solution}

\begin{example}
Find the derivative of \[
f(x)=\frac{(x+2)(x^{2}+1)}{(x+3)(x^{2}-2)}.\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find the derivative of \[
f(x)=\frac{x}{x^{2}+1}.\]

\end{xca}

\subsubsection*{Answer }

\( f'(x)= \) 

\begin{xca}
Find the derivative of \[
g(x)=\frac{x^{2}+3x-2}{(x^{2}+2)(x^{2}+2x+5)}.\]

\end{xca}

\subsubsection*{Answer}

\( g'(x)= \)

\begin{note}
Submit your answers to each group of exercises on this page separately. This
button only sends your answers to these two exercises.
\end{note}

\subsection{The Chain Rule }

The problem here is to find the derivative of a \textit{composition}, \( f(g(x)) \).
The way you should think of the rule is to work ``from the outside in''. We'll
go over a number of examples, after we have the theorem. 

\begin{thm*}
\textbf{\textup{{[}The Chain Rule{]}}}  \( \left( f(g(x))\right) '=f'(g(x))\cdot g'(x). \)
\end{thm*}
This looks a little awkward. It's actually better in Leibniz notation: set \( u=g(x) \),
and \( y=f(u) \). Then, we are asked to find \[
\displaystyle \frac{d}{dx}\left( f(g(x))\right) =\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.\]
  

\begin{proof}
~
\end{proof}
One way of thinking about this is that, since \( dy/dx \) is the rate of change
of \( y \) with respect to change in \( x \), it can be interpreted as the
magnification factor of change. If \( x \) changes by a (small) amount \( \Delta x \),
then \( y \) changes by \( dy/dx \) times as much. Think of a magnifying glass.
Each object is magnified in size by a fixed amount. The chain rule then expresses
the idea that, of you magnify twice in succession (two lenses, one looking through
the other one), the resulting magnification is the product of each of the separate
lenses. 

\begin{example}
Find \[
\left( \sqrt{x^{2}+x}\right) '.\]

\end{example}

\subsubsection*{Solution}

\begin{example}
If \( f(x)=(x^{3}+4x^{2}-3x+1)^{4} \), find \( f'(x) \).
\end{example}

\subsubsection*{Solution}

\begin{xca}
If \[
f(x)=\left( (x^{2}+3x+1)(2x-1)\right) ^{1/3},\]
find \( f'(x) \).
\end{xca}

\subsubsection*{Answer}

\( f'(x)= \)

\begin{xca}
Find the derivative of \[
g(x)=\sqrt{\frac{3x+1}{2x-1}}.\]

\end{xca}

\subsubsection*{Answer}

\( g'(x)= \)

\begin{note}
Submit your answers to each group of exercises on this page separately. This
button only sends your answers to these two exercises.\end{note}

\end{document}
