\begin{frame}
  \frametitle{Differentiation Rules: Chain Rule}

  \begin{block}{}
    \begin{malign}
      (f\circ g)'(x) \;=\; f'(g(x)) \cdot g'(x)
    \end{malign}
  \end{block}  

  \begin{exampleblock}{}
    Let $f(x) = \sqrt{x^2 + 1}$. Find $f'(x)$.
    \pause\medskip
    
    We have that
    \begin{talign}
      f(x) &= g(h(x)) &\text{where}&&
      g(x) &= \sqrt{x} &
      h(x) &= x^2 + 1
    \end{talign}
    \pause
    and
    \begin{talign}
      g'(x) = \frac{1}{2\sqrt{x}} &&
      \mpause[1]{h'(x) = 2x}
    \end{talign}
    \pause\pause
    
    Hence:
    \begin{talign}
      f'(x) = (g\circ h)'(x) = 
      \mpause[1]{ \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x }
      \mpause[2]{= \frac{x}{\sqrt{x^2 + 1}} }
    \end{talign}
  \end{exampleblock}
  \vspace{10cm}
\end{frame}