\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}