Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 189: 37

Tangent line: $y = - \frac{1}{3}x + \frac{5}{6}$ Normal line: $y = 3x - \frac{5}{2}$

$$\eqalign{ & y = \frac{{3x}}{{1 + 5{x^2}}}{\text{ at the point }}\left( {1,\frac{1}{2}} \right) \cr & {\text{Differentiate }}y{\text{ to calculate the slope at the point }}\left( {1,\frac{1}{2}} \right) \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{3x}}{{1 + 5{x^2}}}} \right] \cr & {\text{Use the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + 5{x^2}} \right)\frac{d}{{dx}}\left[ {3x} \right] - 3x\frac{d}{{dx}}\left[ {1 + 5{x^2}} \right]}}{{{{\left( {1 + 5{x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + 5{x^2}} \right)\left( 3 \right) - 3x\left( {10x} \right)}}{{{{\left( {1 + 5{x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{3 + 15{x^2} - 30{x^2}}}{{{{\left( {1 + 5{x^2}} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{3 - 15{x^2}}}{{{{\left( {1 + 5{x^2}} \right)}^2}}} \cr & {\text{Find }}m{\text{ at }}\left( {1,\frac{1}{2}} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 1}} = \frac{{3 - 15{{\left( 1 \right)}^2}}}{{{{\left( {1 + 5{{\left( 1 \right)}^2}} \right)}^2}}} \cr & m = - \frac{1}{3} \cr & \cr & {\text{Use the Point}} - {\text{Slope Form of the Equation of a Line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & \underbrace {\left( {1,\frac{1}{2}} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 1{\text{ and }}{y_1} = \frac{1}{2} \cr & {\text{Therefore}} \cr & y - \frac{1}{2} = - \frac{1}{3}\left( {x - 1} \right) \cr & {\text{Simplify}} \cr & y - \frac{1}{2} = - \frac{1}{3}x + \frac{1}{3} \cr & y = - \frac{1}{3}x + \frac{5}{6} \cr & \cr & {\text{The equation of the normal line is given by}} \cr & y - {y_1} = - \frac{1}{m}\left( {x - {x_1}} \right) \cr & y - \frac{1}{2} = - \frac{1}{{ - 1/3}}\left( {x - 1} \right) \cr & y - \frac{1}{2} = 3\left( {x - 1} \right) \cr & y = 3x - 3 + \frac{1}{2} \cr & y = 3x - \frac{5}{2} \cr & \cr & {\text{Summary}} \cr & {\text{equation of the tangent line: }}y = - \frac{1}{3}x + \frac{5}{6} \cr & {\text{equation of the normal line: }}y = 3x - \frac{5}{2} \cr} $$