\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}