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\begin{frame}
  \frametitle{Review - Midterm Exam 3}

  \begin{exampleblock}{}
    Find $\frac{dy}{dx}$ for the curve
    \begin{talign}
      \sin(x+y) = y^2\cos x
    \end{talign}
    \pause
    We use implicit differentiation:\pause
    \begin{talign}
      &\frac{d}{dx}(\sin(x+y)) = \frac{d}{dx} (y^2\cos x) \\
      &\mpause[1]{\implies \cos(x+y)(1+y') = 2yy'\cos(x) + y^2(-\sin x) } \\
      &\mpause{\implies y'\cos(x+y)  - 2yy'\cos(x) =  -y^2\sin x - \cos(x+y) } \\
      &\mpause{\implies y'(\cos(x+y)  - 2y\cos(x)) =  -y^2\sin x - \cos(x+y) } \\
      &\mpause{\implies \frac{dy}{dx} = y' =  \frac{y^2\sin x + \cos(x+y)}{2y\cos(x)-\cos(x+y)} } \\
    \end{talign}
  \end{exampleblock}
\end{frame}