Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 211: 53

$$y = {e^{ - \frac{2}{e}}}{\text{ and }}y = {e^{\frac{2}{e}}}$$

$$\eqalign{ & y = {\left( {{x^2}} \right)^x} \cr & y = {x^{2x}} \cr & {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr & f\left( x \right) = {e^{2x\ln x}} \cr & {\text{Differentiating}} \cr & f'\left( x \right) = {e^{2x\ln x}}\frac{d}{{dx}}\left[ {2x\ln x} \right] \cr & {\text{Use the product rule}} \cr & f'\left( x \right) = {x^{2x}}\left[ {2x\left( {\frac{1}{x}} \right) + \ln x\left( 2 \right)} \right] \cr & {\text{Multiply}} \cr & f'\left( x \right) = {x^{2x}}\left( {2 + \ln {x^2}} \right) \cr & \cr & {\text{Set the derivative equal to 0}} \cr & {x^{2x}}\left( {2 + \ln {x^2}} \right) = 0 \cr & {x^{2x}} = 0{\text{ or }}2 + \ln {x^2} = 0 \cr & {x^{2x}}{\text{ is always positive with }}x > 0,{\text{ then}} \cr & 2 + \ln {x^2} = 0 \cr & \ln {x^2} = - 2 \cr & {e^{\ln {x^2}}} = {e^{ - 2}} \cr & {x^2} = {e^{ - 2}} \cr & x = \pm \sqrt {{e^{ - 2}}} \cr & x = \pm {e^{ - 1}} \cr & \cr & {\text{Then the function has a horizontal line at the points}} \cr & f\left( {{e^{ - 1}}} \right) = {\left( {{e^{ - 1}}} \right)^{2\left( {{e^{ - 1}}} \right)}} = {e^{ - \frac{2}{e}}} \cr & f\left( { - {e^{ - 1}}} \right) = {\left( { - {e^{ - 1}}} \right)^{2\left( { - {e^{ - 1}}} \right)}} = {e^{\frac{2}{e}}} \cr & {\text{The horizontal lines are}} \cr & y = {e^{ - \frac{2}{e}}}{\text{ and }}y = {e^{\frac{2}{e}}} \cr} $$