\begin{frame}
  \frametitle{Implicit Differentiation}

  \begin{exampleblock}{}
    Find $y'$ where
    \begin{talign}
      \sin (x+y) = y^2 \cos x
    \end{talign}
    \pause\medskip
    We have:
    \begin{talign}
      \frac{d}{dx} \sin (x+y) &= \frac{d}{dx} (y^2 \cos x) \\
      \mpause[1]{ \cos(x+y)\cdot(1 + y') } &\mpause[1]{= \cos x \cdot \frac{d}{dx} (y^2) + y^2 \frac{d}{dx} (\cos x) } \\
      \mpause[2]{ \cos(x+y) + y'\cos(x+y) } &\mpause[2]{= \cos x \cdot (2y y') + y^2 (-\sin x) } \\
      \mpause[3]{ y'\cos(x+y) - 2y y'\cos x} &\mpause[3]{= -y^2 \sin x - \cos(x+y)} \\
      \mpause[4]{ y'(\cos(x+y) - 2y \cos x)} &\mpause[4]{= -(y^2 \sin x + \cos(x+y)) } \\
      \mpause[5]{ y'} &\mpause[5]{= -\frac{y^2 \sin x + \cos(x+y)}{\cos(x+y) - 2y \cos x} } 
    \end{talign}
  \end{exampleblock}
\end{frame}