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\begin{frame}
\frametitle{Implicit Differentiation}

\begin{exampleblock}{}
Find $y'$ where $x^3 + y^3 = 6xy$.
\only<1>{
\begin{center}
\begin{tikzpicture}[default,baseline=1cm,scale=.47]
{\def\diaborderx{1cm}
\diagram{-6}{6}{-6}{6}{1}}
\diagramannotatez
\begin{scope}[ultra thick]
\draw[cgreen,ultra thick] plot[domain=-0.84:0,samples=100] ({3*\x/(1+pow(\x,3))},{3*pow(\x,2)/(1+pow(\x,3))});
\draw[cgreen,ultra thick] plot[domain=-20:-1.19,samples=100] ({3*\x/(1+pow(\x,3))},{3*pow(\x,2)/(1+pow(\x,3))});
\draw[cgreen,ultra thick] plot[domain=0:20,samples=200] ({3*\x/(1+pow(\x,3))},{3*pow(\x,2)/(1+pow(\x,3))});
\end{scope}
\end{tikzpicture}
\end{center}
\smallskip
}
\only<2->{
\pause\medskip
\begin{talign}
& \frac{d}{dx} (x^3 + y^3) = \frac{d}{dx} 6xy \\[1ex]
& \mpause[1]{ 3x^2 + 3y^2 \cdot y' = \frac{d}{dx} 6xy } \\[1ex]
& \mpause[2]{ 3x^2 + 3y^2 \cdot y' = 6x \frac{d}{dx} y + y \frac{d}{dx} 6x} \\[1ex]
& \mpause[3]{ 3x^2 + 3y^2 \cdot y' = 6x y' + 6y} \hspace{1cm} \mpause[4]{ \text{\textcolor{gray}{we solve for $y'$}} }\\[1ex]
& \mpause[5]{ 3y^2 \cdot y' - 6x y'= + 6y - 3x^2 } \\[1ex]
& \mpause[6]{ y'(3y^2 - 6x)= 6y - 3x^2 } \\[1ex]
& \mpause[7]{ y'= \frac{ 6y - 3x^2}{3y^2 - 6x} } \mpause[8]{= \frac{ 2y - x^2}{y^2 - 2x} }
\end{talign}
}
\end{exampleblock}
\vspace{10cm}
\end{frame}