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\centerline{MATH 135 -- Calculus 1}
\centerline{The Chain Rule and Implicit Differentiation}
\centerline{November 4, 2016}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
As we saw in today's video, we may want to compute tangent lines to curves that 
are more general than graphs $y = f(x)$.  (Equivalently we may want to understand
rates of change of two related quantities when the relationship is more complicated
than saying one is a function of the other.)  For instance, consider the curve defined
by the equation 
\begin{equation}
\label{fol}
x^3 + y^3 - 3xy = 0,
\end{equation}
plotted on the back of this sheet.  Note
that this curve is \emph{not a single graph} $y = f(x)$ because \emph{it fails the
vertical line test(!)}.  However, by focusing on just portions of the curve, we can
see that there are several graphs that lie on the curve.  Each of those defines $y$
\emph{implicitly as a function of $x$} on a portion of the curve.  These are called 
\emph{implicit functions} because we will not have explicit formulas for them.  However
we can do things like compute derivatives because in \eqref{fol}, we can \emph{think of 
$y$ as a function of $x$} and differentiate things like $y^3$ by the chain rule, and 
things like the $xy$ using the product rule.  If we do this we get a new equation:
$$3x^2 + 3y^2 \frac{dy}{dx} - 3x\frac{dy}{dx} - 3y = 0.$$
Solving for $\frac{dy}{dx}$, we get
$$\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}.$$
This process is called \emph{implicit differentiation}.
Then, for instance, if we wanted to find points where the tangent line was
horizontal, we would set $\frac{dy}{dx} = 0$, and get $y = x^2$.  So from \eqref{fol},
$$x^3 + x^6 - 3x^3 = 0 \qquad {\rm or} \qquad x^3(x^3 - 2) = 0.$$
This is satisfied when $x = 0$ and also $x = 2^{1/3} \doteq 1.26$.  The point 
$(2^{1/3},2^{2/3})$ is one such point!
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\noindent
{\it Questions}
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\noindent
\begin{enumerate}
\item[1.]  Find $\frac{dy}{dx}$ by implicit differentiation, then perform
the other indicated calculations (if any):
\begin{enumerate}
\item[(a)]  $3x^2 + 2y^2 = 5$.  Use $\frac{dy}{dx}$ to find the equation 
of the tangent line to this curve (an ellipse) at $(1,1)$.
\item[(b)]  $\sin(xy) = x$.
\item[(c)]  $xy + x^2y^2 = 6$.  Find the equation of the tangent line to this
curve at the point $(2,1)$.  
\end{enumerate}
\item[2.]  Find all the points on the curve defined by $y^2 = x^3 - 3x + 1$
where the tangent line is \emph{horizontal}.  (This curve is shown in Figure
10 on page 173 of our book.)
\end{enumerate}
\vfill\eject
\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{Folium.png}
\end{center}
\caption{The curve $x^3 + y^3 - 3xy = 0$ (the ``folium of Descartes'')}
\end{figure}
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