\documentclass{beamer}
\usepackage{appearance}
\newcommand\reals{\mathbb R}
\begin{document}
%MODULE: Continuity
\pause{First Example}

\noindent
{\dims We demonstrate this using the first example.}\begin{align*}
	\dims y'&\dims=-\frac{x}{y}.
\end{align*}
We illustrate this by computing the slopes of lines tangent to the circle at various points.

The point $\hilite (1,0)$ is on the curve, since $\ec 1^2+0^2=1$.
At this point, \[{\ec { y'}=-\frac{1}{0}},\mathrm{\ corresponding\ to\ a\ vertical\ line}.\]
\begin{figure}[h!]
\begin{center}



\includegraphics[scale=0.3]{figures/tanline_at_right.png}
\end{center}\end{figure}



\lyxframe{Implicit Plots}

\noindent
It isn't always easy to view the graph of a relation.
If you can't separate $\ec y$ and $\ec x$, you can still plot it in SAGE using the \texttt{\ttcolor implicit\_plot()} command.
For example, the graphs of the circles were generated using these commands:

\framebox{\parbox{8.75in}{\Large\texttt{\ttcolor circleplot = implicit\_plot(x\^{}2+y\^{}2-1,(x,-2,2),(y,-2,2),cmap='jet',plot\_points=500)}\\
\texttt{\ttcolor circleplot.show(aspect\_ratio=1)}\\
\texttt{\ttcolor upperplot = implicit\_plot(x\^{}2+y\^{}2-1,(x,-2,2),(y,0,2),cmap='jet',plot\_points=500)}\\
\texttt{\ttcolor lowerplot = implicit\_plot(x\^{}2+y\^{}2-1,(x,-2,2),(y,-2,0),cmap='hsv',plot\_points=500)}\\
\texttt{\ttcolor (upperplot+lowerplot).show(aspect\_ratio=1)}}}

\noindent\emph{(Try it!)}

The command takes as its first argument an expression \texttt{\ttcolor exp} which corresponds to the equation $\ec\mathtt{\ttcolor exp}=0$.
We want to plot $\ec x^2+y^2=1$, which corresponds to $\ec x^2+y^2-1=0$, so $\ec\mathtt{\ttcolor exp}=x^2+y^2-1$.
The second and third arguments are a range of $x$- and $y$-values over which the expression should be graphed.
The third argument corresponds to a color
(the usual colors don't apply in the currect version of SAGE;
we advise you to stick with \texttt{\ttcolor 'jet'} (blue), \texttt{\ttcolor 'hsv'} (red), and \texttt{\ttcolor summer} (green)).
The final argument should be included in order to obtain an accurate graph from SAGE;
if the number is too small, the graph will not be accurate.



\lyxframe{Second Example}

\noindent
We can also differentiate the second example, where it is not possible to isolate $\ec y$.
\begin{align*}
	\ec x^2e^{xy}&{\ec=x+y}.
\end{align*}



\pause{Second Example}

\noindent
We can also differentiate the second example, where it is not possible to isolate $\ec y$.
\begin{align*}
	\dims x^2e^{xy}&{\dims=x+y}\\
	\ec 2xe^{xy}+x^2e^{xy}\cdot\left(y+x{\hilite y'}\right)&{\ec=1+{\hilite y'}}.
\end{align*}
\begin{explanation}Use the Product Rule and the Chain Rule.
We only write $ y'$ when we take the derivative of $ y$.\end{explanation}



\pause{Second Example}

\noindent
We can also differentiate the second example, where it is not possible to isolate $\ec y$.
\begin{align*}
	\dims x^2e^{xy}&{\dims=x+y}\\
	\dims 2xe^{xy}+x^2e^{xy}\cdot\left(y+xy'\right)&\dims=1+y'\\
	\ec 2xe^{xy}+x^2ye^{xy}+x^3e^{xy}\cdot {\hilite y'}&{\ec=1+{\hilite y'}}.
\end{align*}
\begin{explanation}Distribution.\end{explanation}



\pause{Second Example}

\noindent
We can also differentiate the second example, where it is not possible to isolate $\ec y$.
\begin{align*}
	\dims x^2e^{xy}&{\dims=x+y}\\
	\dims 2xe^{xy}+x^2e^{xy}\cdot\left(y+xy'\right)&\dims=1+y'\\
	 2xe^{xy}+x^2ye^{xy}+x^3e^{xy}\cdot { y'}&=1+{ y'}\\
	\ec \left(x^3e^{xy}-1\right){\hilite y'}&{\ec=1-e^{xy}\left(x^2+2x\right)}.
\end{align*}
\begin{explanation}Everything with $ y'$ on one side;\\everything without it to the other.\end{explanation}



\pause{Second Example}

\noindent
We can also differentiate the second example, where it is not possible to isolate $\ec y$.
\begin{align*}
	\dims x^2e^{xy}&{\dims=x+y}\\
	\dims 2xe^{xy}+x^2e^{xy}\cdot\left(y+xy'\right)&\dims=1+y'\\
	 2xe^{xy}+x^2ye^{xy}+x^3e^{xy}\cdot { y'}&=1+{ y'}\\
	 \left(x^3e^{xy}-1\right){ y'}&{=1-e^{xy}\left(x^2+2x\right)}\\
	\ec { y'}&{\ec=\frac{1-e^{xy}\left(x^2+2x\right)}{x^3e^{xy}-1}}.
\end{align*}
\begin{explanation}Solve for $ y'$.\end{explanation}



\lyxframe{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\ec y&{\ec=\arcsin x}.
\end{align*}



\pause{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\dims y&{\dims=\arcsin x}\\
	\ec\sin y&{\ec=x}.
\end{align*}
\begin{explanation}Properties of inverse functions.\end{explanation}



\pause{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\dims\sin y&{\dims=x}\\
	\ec\cos y \cdot y'&{\ec=1}.
\end{align*}
\begin{explanation}Differentiated both sides with respect to $x$.\end{explanation}



\pause{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\dims\cos y\cdot y'&{\dims=1}\\
	\ec y'&{\ec=\frac{1}{\cos y}}.
\end{align*}
\begin{explanation}Solve for $ y'$.\end{explanation}



\pause{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\dims y'&{\dims=\frac{1}{\cos y}}\\
	\ec y'&{\ec=\frac{1}{\cos\left(\arcsin x\right)}}.
\end{align*}
\begin{explanation}Substitute $y=\arcsin x$.\end{explanation}



\pause{Third Example}

\noindent
The third example asks us to find the derivatives of the inverse functions of $\ec\sin x$,
$\ec\tan x$, and $\ec\sec x$. The derivatives of these functions have an interesting form.
Implicit differentation allows us to find these derivatives. Start with $\ec\arcsin x$:.
\begin{align*}
	\dims y'&{\dims=\frac{1}{\cos\left(\arcsin x\right)}}\\
	\ec y'&{\ec=\frac{1}{\sqrt{1-x^2}}}.
\end{align*}
\begin{explanation}This is a property learned in precalculus; we explain it briefly.
We want to know $\cos\alpha$ where $\alpha=\arcsin x$.
Thus $x=\sin\alpha$, so we can draw a right triangle with hypotenuse of length $1$
that has an angle of measure $\alpha$ and a leg opposite $\alpha$ of length $x$.
The length of the side adjacent to $\alpha$ will then be $\sqrt{1-x^2}$.
\end{explanation}

\lyxframe{Third Example}

\noindent
We have the following.
\begin{theorem}[(Derivatives of inverse trig functions)]
\begin{itemize}
\item If $\ec f(x)=\arcsin x$, then $\ec f'(x) = \frac{1}{\sqrt{1-x^2}}$.
\item If $\ec f(x)=\arctan x$, then $\ec f'(x) = \frac{1}{{1+x^2}}$.
\item If $\ec f(x)=\arcsec x$, then $\ec f'(x) = \frac{1}{x\sqrt{x^2-1}}$.
\end{itemize}
\end{theorem}
\begin{explanation}You should check on your own that the other two are correct.\end{explanation}



\lyxframe{Conclusion}

\begin{itemize}
\item \emph{\hilite Implicit differentiation} allows us to compute the derivative in cases where $\ec y$ may not be
	a function of $\ec x$.
\item Implicit differentation uses the chain rule: whenever you have to take the derivative of $\ec y$,
you multiply by $\hilite y'$.
\item Implicit differentiation allowed us to determine the derivatives of three new functions:
\begin{theorem}[(Derivatives of inverse trig functions)]
\begin{itemize}
\item If $\ec f(x)=\arcsin x$, then $\ec f'(x) = \frac{1}{\sqrt{1-x^2}}$.
\item If $\ec f(x)=\arctan x$, then $\ec f'(x) = \frac{1}{{1+x^2}}$.
\item If $\ec f(x)=\arcsec x$, then $\ec f'(x) = \frac{1}{x\sqrt{x^2-1}}$.
\end{itemize}
\end{theorem}
You should remember these derivatives.
\end{itemize}
\lyxframeend{}
\end{document}
