\begin{frame}
  \frametitle{2nd Midterm Exam - Review}

  \begin{exampleblock}{}
    Find an equation for the tangent at point $(3,-2)$ to the curve\vspace{-1.2ex}
    \begin{minipage}{.59\textwidth}
    \begin{align*}
      y^2(y^2-4) = x^2(x^2-9)
    \end{align*}
    \onslide<2->{
    We have\vspace{-1.2ex}
    \begin{align*}
      y^4 - 4y^2 = x^4-9x^2
    \end{align*}
    }
    \end{minipage}
    \begin{minipage}{.39\textwidth}
      \vspace{2ex}
      \includegraphics[scale=.6]{../devil.png}
      \vspace{-2ex}
    \end{minipage}\vspace{-2ex}
    \pause\pause
    We use implicit differentiation\vspace{-1.2ex}
    \begin{align*}
      &\frac{d}{dx} (y^4 - 4y^2) = \frac{d}{dx} (x^4-9x^2)\\
      &\mpause[1]{\implies\; 4y^3y' - 8yy' = 4x^3-18x}\\
      &\mpause[2]{\implies\; y'(4y^3 - 8y) = 4x^3-18x}\\
      &\mpause[3]{\implies\; y' = \frac{2x^3-9x}{2y^3 - 4y}}
      \mpause[4]{ = \frac{2\cdot 3^3-9\cdot 3}{2(-2)^3 - 4\cdot (-2)}}
      \mpause[5]{ = \frac{54-27}{-16 + 8}}
      \mpause[6]{ = -\frac{27}{8}}
    \end{align*}
    \pause\pause\pause\pause\pause\pause\pause
    Thus the equation for the tangent is \quad \alert{$y + 2 = -\frac{27}{8}\cdot (x-3)$}.
  \end{exampleblock}
\end{frame}