\begin{frame}
  \frametitle{Implicit Differentiation}

  \begin{exampleblock}{}
    Find $y''$ where
    \begin{talign}
      x^4 + y^4 = 16
    \end{talign}
    \pause\medskip
    We have:
    \begin{talign}
      \frac{d}{dx} (x^4 + y^4) = \frac{d}{dx} 16
      \mpause[1]{\;\;\implies\;\; 4x^3 + 4y^3y' = 0}
      \mpause[2]{\;\;\implies\;\; y' = -\frac{x^3}{y^3}}
    \end{talign}
    \pause\pause\pause
    
    Thus
    \begin{talign}
      y'' &= \frac{d}{dx}\left( -\frac{x^3}{y^3} \right)
      \mpause[1]{ = -\frac{y^3 \frac{d}{dx} x^3  - x^3 \frac{d}{dx} y^3}{(y^3)^2} } 
      \mpause[2]{ = -\frac{y^3 3x^2  - x^3 3y^2y'}{y^6} } \\
      &\mpause[3]{ = -\frac{3x^2y^3  - 3x^3y^2 \left( -\frac{x^3}{y^3} \right) }{y^6} } 
      \mpause[4]{ = -\frac{ 3x^2(x^4 + y^4) }{y^7} } 
      \mpause[5]{ = -\frac{ 48x^2 }{y^7} } 
    \end{talign}
  \end{exampleblock}
\end{frame}