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

  \begin{exampleblock}{}
    Find all points on the curve where the slope of the tangent is $-1$\vspace{-1.2ex}
    \begin{align*}
      x^2y^2 + xy = 2
    \end{align*}\vspace{-3.5ex}
    \pause
    
    \only<-6>{
    We use implicit derivatives:\vspace{-1.2ex}
    \begin{align*}
      &\frac{d}{dx}(x^2y^2 + xy) = \frac{d}{dx} 2\\
      &\mpause[1]{\implies\; x^2 \frac{d}{dx}y^2 + y^2 \frac{d}{dx}x^2 \quad+\quad x\frac{d}{dx}y + y \frac{d}{dx}x = 0 }\\
      &\mpause[2]{\implies\; x^2 2yy' + y^2 2x \quad+\quad xy' + y = 0 }\\
      &\mpause[3]{\stackrel{y'=-1}{\implies}\; -2x^2y + 2xy^2 - x + y = 0 }\\
      &\mpause[4]{\implies\; (-x+y) \cdot (2xy+1) = 0 }
    \end{align*}\vspace{-3.5ex}
    }
    \pause\pause\pause\pause\pause
    
    The slope is $-1$ if \quad$x = y$\quad or \quad$xy = -1/2$\quad.\\\pause
    \only<7->{   
    \medskip\pause
     
    But when is $(x,y)$ on the original curve?\pause\vspace{-1.2ex}
    \begin{align*}
      x= y &\mpause[1]{\;\implies\; (x^2)^2 + x^2 = 2} \\
           &\mpause[2]{\;\implies\; (x^2-1)(x^2+ 2) = 0} \\
           &\mpause[3]{\;\implies\; x^2 = 1} \\
           &\mpause[4]{\;\implies\; x = \pm 1} \\
           &\mpause[5]{\;\implies\; \text{on the curve if $x = y = \pm 1$}}\\
      \mpause[6]{xy= -1/2} 
           &\mpause[7]{\;\implies\; (xy)^2 + xy  2} \\ 
           &\mpause[8]{\;\implies\; 1/4 + 1/2 = 2} \\
           &\mpause[9]{\;\implies\; \text{can never be on the curve}}
    \end{align*}\vspace{-3ex}
    }
    \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause
    
    The points on the curve with slope $-1$ are $(1,1)$ and $(-1,-1)$.
  \end{exampleblock}
  \vspace{10cm}
\end{frame}