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\begin{document}

\title{Math 241, Lecture 17}
\maketitle

\subsection{ Some practice with logarithms, exponential functions, and
the chain rule}  

Derivative rules are great fun because they can be combined.  
Taking the derivative of some functions requires every rule we know. 

\begin{example}
Take the derivatives of $(\frac{e^x \ln x}{x^2 + 1})^9$ and
$\ln (\frac{e^x}{\sqrt{x}})^9$.  
\end{example}

\section{ Implicit Differentiation }

At the beginning of the term, we discussed the usefulness of functions
in describing many different relationships.   But we also emphasized that not 
all relationships are described by functions.    For example, in economics
supply and demand are related, but one is not a function of the other.  
In mathematics, the equation $x^2 + y^2 = 25$ is one of the most basic;  
the points which satisfy it constitute a circle of radius $5$.  

Amazingly, we can follow rules for taking derivatives to find the rate
of change of one variable with respect to another in some cases
where there is a relationship between them which is not  described by a function.  

It is best to start with examples and then summarize the general method as and
after we do them.  

\begin{example}
Find the equation of the tangent line to the circle $x^2 + y^2 = 25$ at the 
point $(3,4)$.  
\end{example}

Here are the steps:  
\begin{itemize}
\item Remember the point-slope form for a line, reducing problem to
finding $\frac{dy}{dx}$.  
\item Differentiate the equation, treating $y$ as a function of $x$ even though
it is not expressed as one.    Remember to use the chain rule to differentiate  $y$,
since we are treating it as a function of $x$!  
\item The resulting expression has $\frac{dy}{dx}$ in it, since we have used the chain
rule on $y$.    Now do algebra to solve for it.  
\item Now that we know a formula for $\frac{dy}{dx}$, we can plug in values for
$x$ and $y$ to find the slope of the tangent line and finish the problem.  			
\end{itemize}

Note that we could have done this problem by solving for $y$,
namely $y = \sqrt{25 - x^2}$, and finding the slope by taking the derivative as we are
used to.    This is a good way to check that our answer is correct in the previous case.  

  

We practice with a couple examples before trying to apply this technique to more 
complicated problems.  

\begin{example}
\begin{itemize}
\item Find $\frac{du}{dv}$ if $u$ and $v$ are related by $u^3 v - 2u^2 v^2 + v^4 = 17$.  
\item Find $\frac{dy}{dx}$ when $e^{xy + 4x -3} = 5x^2$.   (Do this twice, the second
time by taking the natural logarithm of both sides).
\end{itemize}
\end{example}

 

\begin{example}
Two manufacturers of widgets are in direct competition.  Because of the many
variables in pricing and publicity, the number of widgets they sell does
not add up to a constant, but $6x^2 + xy + 5y^2 = 120,000$, where
$x$ is the number of widgets sold per day
by the first company and $y$ by the second.  
If both companies are currently selling $100$ widgets, what would be the
effect on the first company if the second is increase sales by ten widgets per day.  
\end{example}

\end{document}
