Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 189: 38

Tangent line:$y = 2x $ Normal line: $y = - \frac{1}{2}x $

$$\eqalign{ & y = x + x{e^x}{\text{ at the point }}\left( {0,0} \right) \cr & {\text{Differentiate }}y{\text{ to calculate the slope at the point }}\left( {0,0} \right) \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x + x{e^x}} \right] \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ x \right] + \underbrace {\frac{d}{{dx}}\left[ {x{e^x}} \right]}_{{\text{product rule}}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {{e^x}} \right]{\text{ + }}{e^x}\frac{d}{{dx}}\left[ x \right] \cr & \frac{{dy}}{{dx}} = 1 + x{e^x}{\text{ + }}{e^x}\left( 1 \right) \cr & \frac{{dy}}{{dx}} = 1 + x{e^x}{\text{ + }}{e^x} \cr & \cr & {\text{Find }}m{\text{ at }}\left( {0,0} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = 1 + \left( 0 \right){e^0}{\text{ + }}{e^0} \cr & m = 2 \cr & \cr & {\text{Use the Point}} - {\text{Slope Form of the Equation of a Line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & \underbrace {\left( {0,0} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 0{\text{ and }}{y_1} = 0 \cr & {\text{Therefore}} \cr & y - 0 = 2\left( {x - 0} \right) \cr & {\text{Simplify}} \cr & y = 2x \cr & \cr & {\text{The equation of the normal line is given by}} \cr & y - {y_1} = - \frac{1}{m}\left( {x - {x_1}} \right) \cr & y - 0 = - \frac{1}{2}\left( {x - 0} \right) \cr & y = - \frac{1}{2}x \cr & \cr & {\text{Summary}} \cr & {\text{equation of the tangent line: }}y = 2x \cr & {\text{equation of the normal line: }}y = - \frac{1}{2}x \cr} $$