## Calculus with Applications (10th Edition)

$$y = x + 2$$
\eqalign{ & {x^2}{y^2} = 1;{\text{ }}\left( { - 1,1} \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( 1 \right) \cr & {\text{use product rule}} \cr & {x^2}\frac{d}{{dx}}\left( {{y^2}} \right) + {y^2}\frac{d}{{dx}}\left( {{x^2}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & {x^2}\left( {2y} \right)\frac{{dy}}{{dx}} + {y^2}\left( {2x} \right) = 0 \cr & 2{x^2}y\frac{{dy}}{{dx}} + 2x{y^2} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & 2{x^2}y\frac{{dy}}{{dx}} = - 2x{y^2} \cr & \frac{{dy}}{{dx}} = - \frac{{2x{y^2}}}{{2{x^2}y}} \cr & \frac{{dy}}{{dx}} = - \frac{y}{x} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 1,1} \right)}} = - \frac{1}{{ - 1}} = 1 \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 - 1 = x - \left( { - 1} \right) \cr & y - 1 = x + 1 \cr & y = x + 1 + 1 \cr & y = x + 2 \cr}