\begin{frame}
  \frametitle{Logarithmic Differentiation}
  
  The following method is called \emph{logarithmic differentiation}.
  \begin{exampleblock}{}
    Differentiate
    \begin{talign}
      y = \frac{ x^{\frac{3}{4}}\cdot \sqrt{x^2+1} }{ (3x+2)^5 }
    \end{talign}
    \pause
    \only<-6>{%
    We take logarithms on both sides:
    \begin{talign}
      \ln y = \ln \frac{ x^{\frac{3}{4}}\cdot \sqrt{x^2+1} }{ (3x+2)^5 } 
            &= \mpause[1]{\ln x^{\frac{3}{4}} + \ln \sqrt{x^2+1} - \ln(3x+2)^5 }\\
            &\mpause[2]{= \frac{3}{4} \ln x + \frac{1}{2}\ln(x^2+1) - 5\ln(3x+2) }
    \end{talign}
    \pause\pause\pause
    We use implicit differentiation:
    \begin{talign}
      \frac{d}{dx} \ln y &= \frac{3}{4} \frac{d}{dx}\ln x + \frac{1}{2} \frac{d}{dx}\ln(x^2+1) - 5 \frac{d}{dx}\ln(3x+2) \\[1ex]
      \mpause[1]{\frac{1}{y}y'} &\mpause[1]{= \frac{3}{4} \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x^2+1}\cdot 2x - 5\frac{1}{3x+2}\cdot 3 }
    \end{talign}
    }%
    \only<7->{
    \pause[7]%
    We have:
    \begin{talign}
      \frac{1}{y}y' &= \frac{3}{4} \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x^2+1}\cdot 2x - 5\frac{1}{3x+2}\cdot 3
    \end{talign}
    \pause
    
    Solving for $y'$ yields:
    \begin{talign}
      y' = y\left(\frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2}\right)
    \end{talign}
    \pause
    Hence
    \begin{talign}
      y' = \frac{ x^{\frac{3}{4}}\cdot \sqrt{x^2+1} }{ (3x+2)^5 } \cdot \left(\frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2}\right)
    \end{talign}
    }
  \end{exampleblock}
  \vspace{10cm}
\end{frame}