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

  \begin{exampleblock}{}
    Find the derivative of $f(x)$ when $0 < x < 5$:
    \begin{align*}
      f(x) = \frac{|x|}{\sqrt{5^2 - x^2}}
    \end{align*}
    \pause
    If $0 < x$ then $|x| = x$. Then
    \begin{align*}
      f(x) &= \frac{x}{\sqrt{5^2 - x^2}}\\
      \mpause[1]{f'(x)} &\mpause[1]{=} \mpause[2]{\frac{1\cdot \sqrt{5^2 - x^2} - x\cdot \frac{1}{2}(5^2 - x^2)^{-\frac{1}{2}}\cdot (-2x)}{(\sqrt{5^2 - x^2})^2}} \\
      &\mpause[3]{= \frac{(5^2 - x^2) + x^2}{(\sqrt{5^2 - x^2})^3}} 
      \mpause[4]{= \frac{25}{(5^2 - x^2)^{\frac{3}{2}}}}
    \end{align*}
    \pause\pause\pause\pause\pause
    What is the left-hand derivative at $0$? \pause Then $x < 0$, thus $|x| = -x$.
    \pause
    \begin{align*}
      \text{left-hand derivative at $0$} = -1/5
    \end{align*}
  \end{exampleblock}
\end{frame}