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

  \begin{exampleblock}{}
    Find the derivative of
    \begin{align*}
      f(x) = \left(\frac{1}{x^3} - 3\frac{1}{x}\right)\cdot (2x^4+ 5x)
    \end{align*}
    using the product rule. \pause 
    \begin{align*}
      &f'(x) = \mpause[1]{ \left(\frac{1}{x^3} - 3\frac{1}{x}\right) \cdot \frac{d}{dx} (2x^4+ 5x)
        + (2x^4+ 5x) \frac{d}{dx} \left(\frac{1}{x^3} - 3\frac{1}{x}\right) } \\
     &\mpause[2]{=  \left(\frac{1}{x^3} - 3\frac{1}{x}\right) \cdot (8x^3+ 5)
        + (2x^4+ 5x) \left(-3x^{-4} - 3\cdot (-1)x^{-2}\right) }   \\
     &\mpause[3]{=  \left(\frac{1}{x^3} - 3\frac{1}{x}\right) \cdot (8x^3+ 5)
        + (2x^4+ 5x) \left(\frac{-3}{x^{4}} + \frac{3}{x^{2}}\right) } \\   
     &\mpause[4]{=  \left(8 - 24x^2 + \frac{5}{x^3} - \frac{15}{x}\right)
        + \left(-6 + 6x^2 - \frac{15}{x^3} + \frac{15}{x} \right) } \\
     &\mpause[5]{=  2 - 18x^2 - \frac{10}{x^3} }
    \end{align*}
  \end{exampleblock}
\end{frame}