46/155
\begin{frame}
  \frametitle{Differentiation Rules: Product Rule}
  
  \begin{exampleblock}{}
    Let $f(x) = xe^x$. Find $f'(x)$.
    \pause
    \begin{talign}
      f'(x) 
      &\mpause[1]{= \frac{d}{dx} (x \cdot e^x)} \\
      &\mpause[2]{= x \frac{d}{dx} (e^x) + e^x \frac{d}{dx} (x)} \\
      &\mpause[3]{= x e^x + e^x} \\
      &\mpause[4]{= (x+1) e^x} 
    \end{talign}
  \end{exampleblock}
  \pause\pause\pause\pause\pause

  \begin{exampleblock}{}
    Let $f(x) = xe^x$. Find the $n$-th derivative $f^{(n)}(x)$.
    \pause
    \begin{talign}
      f''(x) 
      &\mpause[1]{= \frac{d}{dx} (x e^x + e^x)} 
      \mpause[2]{= (x+1) e^x + e^x} 
      \mpause[3]{= (x+2) e^x} \\
      \mpause[4]{f'''(x)} 
      &\mpause[5]{= \frac{d}{dx} (x e^x + 2e^x)} 
      \mpause[6]{= (x+1) e^x + 2e^x} 
      \mpause[7]{= (x+3) e^x}
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause\pause
    Thus obviously we have
    \begin{talign}
      f^{(n)}(x) 
      &= (x+n) e^x
    \end{talign}
  \end{exampleblock}
\end{frame}