\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}