\begin{frame}
  \frametitle{Logarithmic Differentiation}
  
  \begin{exampleblock}{}
    Differentiate
    \begin{talign}
      y = x^{\sqrt{x}}
    \end{talign}
    \pause
    The power rule does not apply: the exponent contains $x$!
    \pause\medskip
    
    We use logarithmic differentiation:
    \begin{talign}
      &\ln y = \ln \left(x^{\sqrt{x}}\right) = \mpause[1]{\sqrt{x} \cdot \ln x} \\
      &\mpause[2]{ \frac{d}{dx} \ln y = \frac{d}{dx } \left( \sqrt{x} \cdot \ln x \right) }\\
      &\mpause[3]{ \frac{1}{y}y' = \sqrt{x}\cdot \frac{1}{x} + \ln x \cdot \frac{1}{2\sqrt{x}} }\\
      &\mpause[4]{ y' = y\left( \frac{1}{\sqrt{x}} + \frac{\ln x}{2\sqrt{x}} \right) }
      \mpause[5]{= y\left( \frac{2+ \ln x}{2\sqrt{x}} \right) }
      \mpause[6]{= x^{\sqrt{x}} \left( \frac{2+ \ln x}{2\sqrt{x}} \right) }
    \end{talign}
    \pause\pause\pause\pause\pause\pause
  \end{exampleblock}
  \pause\medskip
  
  Alternative:
  \quad $x^{\sqrt{x}} \mpause[1]{ = (e^{\ln x})^{\sqrt{x}}} \mpause[2]{ = e^{\ln x \cdot \sqrt{x}}}$\quad
  \pause\pause\pause now use chain rule
  \bigskip
\end{frame}