## Calculus with Applications (10th Edition)

$$y = \frac{3}{4}x + \frac{7}{4}$$
\eqalign{ & y = \sqrt {6x - 2} ,\,\,\,\,\,x = 3 \cr & {\text{find the derivative of }}y \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {6x - 2} } \right] \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {6x - 2} \right)}^{1/2}}} \right] \cr & {\text{by using the chain rule}} \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {6x - 2} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {6x - 2} \right] \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {6x - 2} \right)^{ - 1/2}}\left( 6 \right) \cr & \frac{{dy}}{{dx}} = \frac{3}{{{{\left( {6x - 2} \right)}^{1/2}}}} \cr & \frac{{dy}}{{dx}} = \frac{3}{{\sqrt {6x - 2} }} \cr & {\text{Find the slope of the tangent line at }}x = 3 \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 3}} \cr & m = \frac{3}{{\sqrt {6\left( 3 \right) - 2} }} \cr & m = \frac{3}{4} \cr & \cr & {\text{Evaluate the function at }}x = 3 \cr & y\left( 3 \right) = \sqrt {6\left( 3 \right) - 2} \cr & y\left( 3 \right) = 4 \cr & {\text{we know the point }}\left( {3,4} \right){\text{ and the slope }}m = \frac{3}{4} \cr & {\text{find the equation of the tangent line using the point - slope form of a line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 4 = \frac{3}{4}\left( {x - 3} \right) \cr & {\text{simplifying}} \cr & y - 4 = \frac{3}{4}x - \frac{9}{4} \cr & y = \frac{3}{4}x - \frac{9}{4} + 4 \cr & y = \frac{3}{4}x + \frac{7}{4} \cr}