\begin{frame}
  \frametitle{Review - Midterm Exam 3}

  \begin{exampleblock}{}
    Consider the curve:
    \begin{talign}
      x^2 + y^2 = 1
    \end{talign}
    At what point in the first quadrant has the curve slope $-1$?
    \pause\medskip
    
    We use implicit differentiation:
    \begin{talign}
      \frac{d}{dx}(x^2 + y^2) &= \frac{d}{dx} 1\\
      \mpause[1]{2x + 2yy' &= 0} \\
      \mpause{y' &= -\frac{x}{y}}
    \end{talign}\vspace{-2.25ex}
    \pause\pause\pause
    
    We know $x^2 + y^2 = 1$ and $y > 0$, thus $y = \pause \sqrt{1-x^2}$.
    \pause
    \begin{talign}
      -1 &= y' = -\frac{x}{\sqrt{1-x^2}}
      \mpause[1]{\;\implies\; \sqrt{1-x^2} = x}
      \mpause{\;\implies\; 1-x^2 = x^2}\\
      \mpause{&\implies\; 2x^2 = 1}
      \mpause{\;\implies\; x = \sqrt{1/2}}
    \end{talign}
    \pause\pause\pause\pause
    Thus the slope is $-1$ at point $(\sqrt{1/2},\sqrt{1/2})$.
  \end{exampleblock}  
\end{frame}