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\title{Math 241, Lecture 16}
\maketitle


\subsection{Partial substitution in applications of the chain rule}  

When we are using the chain rule to get a numerical
answer,   as opposed to a formula for the derivative,  
we can save a small amount of work by not substituting
fully to get a closed expression.      The ``intermediate form''
is good enough to plug in and get numerical answers.  

 

\begin{example}
A shoe maker estimates that the profit for selling shoes as
a function of price (accounting for the market equilibrium)
is $-5 + {p} - \frac{p^2}{40}$ per shoe.    [Check that this makes sense.]  
Because of a price war, the shoe maker estimates that the
price will be $22-\sqrt{t}$ dollars over the next $t$ months. How
fast will his profits changing over in $4$ and $9$ months?  
\end{example}
 


\section{The derivative of exponential functions}

Looking at the definition of the derivative of $a^x$, we have
$$\lim_{h \to 0} \frac{a^{x + h} - a^{x}}{h}   = \lim_{h \to 0} \frac{a^x[x^h - 1]}{h}.   $$
But notice that $a^x$ has no dependence on $h$ so we can pull it out of the limit
and get $a^x \lim_{h \to 0} \frac{a^h - 1}{h}$.     This last limit is going to
be some constant, independent of $x$.    It turns out to be equal to
$\ln(a)$!   

\begin{theorem}
The derivative of $a^x$ = $\ln(a) a^x$.    In particular $\frac{d}{dx} e^x  = e^x$.  
\end{theorem}

It is remarkable that the derivative of $e^x$ is itself.   This is a special property
of exponentiation and the number $e$.    

When we compute a derivative, we should see how our computation fits
with the graph of the function.     In the case of $a^x$, we see that as this function
gets larger it grows faster (and thus gets even larger and grows even faster...).  

Knowing just the derivative of $e^x$ allows us to compute the derivatives of functions
made from $e^x$ and polynomials using the chain, product and quotient rules.  

\begin{example}
Find the derivatives of  
\begin{itemize}
\item $e^{x^2}$  
\item $10^x$  
\item $\frac{e^x}{x}$  
\item $A(t) = P e^{rt}$  
\end{itemize}
\end{example}

The fact that the derivative of $e^{rx}$ is $r$ times itself makes it useful in 
modeling populations, investments, temperatures... 

 

\section{The derivative of the natural logarithm function}  

Recall from the properties of logarithm functions that $\log_a(x) = \log_a e  \times  \ln x$.  
We differentiate $\ln x$, since all other logarithm functions differ from it by a constant.  

\begin{theorem}
The derivative of $\ln(x)$ is $\frac{1}{x}$.  
\end{theorem}

We will establish this fact once we learn about {\em implicit differertiation}.  

Note that this fills in a spot which has been missing on the list of derivatives.  
In general the derivative of $\frac{1}{n+1} x^{n+1} = x^n$.    But this does not work 
for $n = -1$.    But in the function $\ln(x)$ has derivative $x^{-1}$.  

Again, we should check the behavior of the derivative with the behavior of the
graph.  

And again, once we combine with other rules, we can now go wild taking derivatives 
of complicated functions.  

\begin{example}
Find the derivatives of the functions
\begin{itemize}
\item  $f(x) = x \ln(x)$  
\item $g(x) = \ln (x^{\frac{7}{2}} + 3)$  
\item $u(t) = \ln x^2$  
\item $(\frac{e^x \ln x}{x^2 + 1})^9$  
\end{itemize}
\end{example}



\end{document}
 