## Calculus with Applications (10th Edition)

$$y = \frac{4}{3}x - \frac{{50}}{3}$$
\eqalign{ & {x^2} + {y^2} = 100;{\text{ }}\left( {8, - 6} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {100} \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( {100} \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & 2x + 2y\frac{{dy}}{{dx}} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & 2y\frac{{dy}}{{dx}} = - 2x \cr & \frac{{dy}}{{dx}} = - \frac{x}{y} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {8, - 6} \right)}} = - \frac{8}{{ - 6}} = \frac{4}{3} \cr & {\text{find the equation of the tangent line at the point }}\left( {8, - 6} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 6 = \frac{4}{3}\left( {x - 8} \right) \cr & y + 6 = \frac{4}{3}x - \frac{{32}}{3} \cr & y = \frac{4}{3}x - \frac{{32}}{3} - 6 \cr & y = \frac{4}{3}x - \frac{{50}}{3} \cr}