\begin{frame} \frametitle{2nd Midterm Exam - Review} \begin{exampleblock}{} Find the derivative of \begin{talign} y = \sqrt{x}\cdot (1+x^2)^{\sin(x)} \end{talign}\vspace{-2.5ex} \pause We have: \begin{talign} \ln y &= \ln \left[ \sqrt{x}\cdot (1+x^2)^{\sin(x)} \right] \mpause[1]{= \ln \sqrt{x} + \ln (1+x^2)^{\sin(x)}} \\ &\mpause[2]{= \frac{1}{2}\cdot \ln x + \sin(x) \cdot \ln (1+x^2)} \end{talign}\vspace{-3ex} \pause\pause\pause Thus \begin{talign} &\frac{d}{dx} \ln y = \frac{d}{dx} \left[ \frac{1}{2}\cdot \ln x + \sin(x) \cdot \ln (1+x^2) \right] \\ &\mpause[1]{\frac{y'}{y} = \left[ \frac{1}{2x} + \sin(x) \cdot \frac{d}{dx}\ln (1+x^2) + \ln (1+x^2) \cdot \frac{d}{dx} \sin(x) \right] } \\ &\mpause[2]{y' = y\left[ \frac{1}{2x} + \sin(x) \frac{1}{1+x^2}2x + (\ln (1+x^2)) \cdot \cos(x) \right] } \\ &\mpause[3]{y' = \left( \sqrt{x}\cdot (1+x^2)^{\sin(x)} \right) \left[ \frac{1}{2x} + \frac{2x\sin(x)}{1+x^2} + \cos(x)\ln (1+x^2) \right] } \\ \end{talign} \end{exampleblock} \end{frame}