Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 202: 88

\[\begin{align} & \text{No horizontal tangent line} \\ & \text{Vertical tangent lines at}\left( -e\sqrt{e},3 \right),\left( e\sqrt{e},3 \right) \\ \end{align}\]

\[\begin{align} & {{x}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & \text{Differentiate implicitly with respect to }x \\ & \frac{d}{dx}\left[ {{x}^{2}}\left( y-2 \right)-{{e}^{y}} \right]=\frac{d}{dx}\left[ 0 \right] \\ & {{x}^{2}}\left( \frac{dy}{dx}-0 \right)+2x\left( y-2 \right)-{{e}^{y}}\frac{dy}{dx}=0 \\ & \text{Solve for }\frac{dy}{dx} \\ & {{x}^{2}}\frac{dy}{dx}+2x\left( y-2 \right)-{{e}^{y}}\frac{dy}{dx}=0 \\ & {{x}^{2}}\frac{dy}{dx}-{{e}^{y}}\frac{dy}{dx}=2x\left( 2-y \right) \\ & \left( {{x}^{2}}-{{e}^{y}} \right)\frac{dy}{dx}=2x\left( 2-y \right) \\ & \frac{dy}{dx}=\frac{2x\left( 2-y \right)}{{{x}^{2}}-{{e}^{y}}} \\ & \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{2x\left( 2-y \right)}{{{x}^{2}}-{{e}^{y}}}=0 \\ & 2x\left( 2-y \right)=0 \\ & x=0,\text{ }y=2 \\ & \text{Substitute }x=0\text{ into }{{x}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & {{\left( 0 \right)}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & -{{e}^{y}}=0 \\ & {{e}^{y}}=0 \\ & \text{Where }{{e}^{y}}\text{ is always positive}\text{.} \\ & \text{Substitute }y=2\text{ into }{{x}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & \text{ }{{x}^{2}}\left( 2-2 \right)-{{e}^{2}}=0 \\ & -{{e}^{2}}\ne 0 \\ & \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{2x\left( 2-y \right)}{{{x}^{2}}-{{e}^{y}}}=\infty \\ & {{x}^{2}}-{{e}^{y}}=0 \\ & {{x}^{2}}={{e}^{y}} \\ & \\ & \text{Substituting }{{e}^{y}}\text{ for }{{x}^{2}}\text{ into }{{x}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & {{e}^{y}}\left( y-2 \right)-{{e}^{y}}=0 \\ & {{e}^{y}}\left( y-2-1 \right)=0 \\ & {{e}^{y}}\left( y-3 \right)=0 \\ & y=3 \\ & \text{Substituting }y=3\text{ into }{{x}^{2}}\left( y-2 \right)-{{e}^{y}}=0 \\ & {{x}^{2}}\left( 3-2 \right)-{{e}^{3}}=0 \\ & {{x}^{2}}={{e}^{3}} \\ & x=\pm \sqrt{{{e}^{3}}} \\ & x=\pm e\sqrt{e} \\ & \text{We obtain the points }\left( -e\sqrt{e},3 \right)\text{ and }\left( e\sqrt{e},3 \right) \\ & \text{The points at which the curve has slope }\infty \text{ are }\left( -e\sqrt{e},3 \right),\left( e\sqrt{e},3 \right) \\ & \\ & \text{This graph confirms the result} \\ \end{align}\]