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\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}