Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 202: 86

\[\begin{align} & \text{Does not exist points at which the curve has slope }0 \\ & \text{Does not exist points at which the curve has slope }\infty \\ \end{align}\]

\[\begin{align} & {{y}^{2}}-3xy=2 \\ & \text{Differentiate implicitly with respect to }x \\ & \frac{d}{dx}\left[ {{y}^{2}}-3xy \right]=\frac{d}{dx}\left[ 2 \right] \\ & 2y\frac{dy}{dx}-3x\frac{dy}{dx}-3y=0 \\ & \text{Solve for }\frac{dy}{dx} \\ & 2y\frac{dy}{dx}-3x\frac{dy}{dx}=3y \\ & \left( 2y-3x \right)\frac{dy}{dx}=3y \\ & \frac{dy}{dx}=\frac{3y}{2y-3x} \\ & \text{The points on the curve at which the tangent line is horizontal } \\ & \text{are when the slope is 0, then} \\ & \frac{dy}{dx}=\frac{3y}{2y-3x}=0 \\ & 3y=0 \\ & y=0 \\ & \text{Substitute }y=0\text{ into }{{y}^{2}}-3xy=2\text{ to find }x \\ & {{0}^{2}}-3x\left( 0 \right)=2 \\ & 0\ne 2 \\ & \text{Therefore, does not exist points at which the curve has slope 0} \\ & \\ & \text{The points on the curve at which the tangent line is vertical} \\ & \text{are when the slope is }\infty \text{, then} \\ & \frac{dy}{dx}=\frac{3y}{2y-3x}=\infty \\ & \frac{3y}{2y-3x}=\infty \\ & 2y-3x=0 \\ & 2y=3x \\ & y=\frac{3x}{2} \\ & \text{Substituting }y=\frac{3x}{2}\text{ into }{{y}^{2}}-3xy=2 \\ & {{\left( \frac{3x}{2} \right)}^{2}}-3x\left( \frac{3x}{2} \right)=2 \\ & \frac{9{{x}^{2}}}{4}-\frac{9{{x}^{2}}}{4}=2 \\ & 0\ne 2 \\ & \text{Does not exist points at which the curve has slope }\infty \\ & \\ & \text{This graph confirms the result} \\ \end{align}\]