\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}  
  \pause
  
  \begin{block}{}
    In general (combining the power and chain rule) we have:
    \begin{talign}
      \frac{d}{dx}[g(x)]^n \;=\; \mpause[1]{ n\cdot[g(x)]^{n-1} \cdot g'(x) }
    \end{talign}
    if $g(x)$ is differentiable.
  \end{block}
  \pause\pause
  
  \begin{exampleblock}{}
    Differentiate
    \begin{talign}
      f(x) = \frac{1}{\sqrt[3]{x^2+ x+ 1}}
    \end{talign}
    \pause
     We have
    \begin{talign}
      f(x) &= (x^2+ x+ 1)^{-\frac{1}{3}}\\
      \mpause[1]{f'(x) }&\mpause[1]{= }\mpause[2]{-\frac{1}{3} \cdot (x^2+ x+ 1)^{-\frac{4}{3}} \cdot (2x+1)}
    \end{talign}
  \end{exampleblock}
  \vspace{10cm}
\end{frame}