## Calculus with Applications (10th Edition)

$$y = \frac{4}{3}x + \frac{{10}}{3}$$
\eqalign{ & {x^2}{y^3} = 8;{\text{ }}\left( { - 1,2} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2}{y^3}} \right) = \frac{d}{{dx}}\left( 8 \right) \cr & {\text{use product rule}} \cr & {x^2}\frac{d}{{dx}}\left( {{y^3}} \right) + {y^3}\frac{d}{{dx}}\left( {{x^2}} \right) = \frac{d}{{dx}}\left( 8 \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & {x^2}\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + {y^3}\left( {2x} \right) = 0 \cr & 3{x^2}{y^2}\frac{{dy}}{{dx}} + 2x{y^3} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & 3{x^2}{y^2}\frac{{dy}}{{dx}} = - 2x{y^3} \cr & \frac{{dy}}{{dx}} = \frac{{ - 2x{y^3}}}{{3{x^2}{y^2}}} \cr & \frac{{dy}}{{dx}} = - \frac{{2y}}{{3x}} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 1,2} \right)}} = - \frac{{2\left( 2 \right)}}{{3\left( { - 1} \right)}} = \frac{4}{3} \cr & {\text{find the equation of the tangent line at the point }}\left( { - 1,1} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = \frac{4}{3}\left( {x - \left( { - 1} \right)} \right) \cr & y - 2 = \frac{4}{3}\left( {x + 1} \right) \cr & y = \frac{4}{3}\left( {x + 1} \right) + 2 \cr & y = \frac{4}{3}x + \frac{4}{3} + 2 \cr & y = \frac{4}{3}x + \frac{{10}}{3} \cr}