\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}{}
  Use 
  \begin{talign}
    \frac{d}{dx} e^x = e^x
  \end{talign}
  and the chain rule to prove
  \begin{talign}
    \frac{d}{dx} a^x = \ln a \cdot a^x
  \end{talign}
  \pause\medskip
  We have
  \begin{talign}
    a^x \;=\; \mpause[1]{(e^{\ln a})^x} \mpause[2]{ = e^{\ln a\cdot x}} 
  \end{talign}
  \pause\pause\pause
  and $f = g\circ h$ where \;\;$g(x) = \mpause[1]{e^x}$\;\; and \;\;$h(x) = \mpause[2]{\ln a \cdot x}$\;\;. 
  \pause\pause\pause Thus
  \begin{talign}
    f'(x) = g'(h(x)) \cdot h'(x) = \mpause[1]{ e^{\ln a \cdot x} \cdot \ln a}
    \mpause[2]{= \ln a \cdot a^x}
  \end{talign}
  \end{exampleblock}
  
  \vspace{10cm}
\end{frame}