## Calculus with Applications (10th Edition)

$$y = 40x - 72$$
\eqalign{ & f\left( x \right) = {x^2}\sqrt {{x^4} - 12} {\text{, }}x = 2 \cr & {\text{find the derivative of }}f\left( x \right) \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2}\sqrt {{x^4} - 12} } \right] \cr & {\text{by using the product rule}} \cr & \frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left[ {\sqrt {{x^4} - 12} } \right] + \sqrt {{x^4} - 12} \frac{d}{{dx}}\left[ {{x^2}} \right] \cr & {\text{find derivatives}} \cr & \frac{{dy}}{{dx}} = {x^2}\left( {\frac{{4{x^3}}}{{2\sqrt {{x^4} - 12} }}} \right) + \sqrt {{x^4} - 12} \left( {2x} \right) \cr & \frac{{dy}}{{dx}} = \frac{{2{x^5}}}{{\sqrt {{x^4} - 12} }} + 2x\sqrt {{x^4} - 12} \cr & {\text{Find the slope of the tangent line at }}x = 2 \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} \cr & m = \frac{{2{{\left( 2 \right)}^5}}}{{\sqrt {{{\left( 2 \right)}^4} - 12} }} + 2\left( 2 \right)\sqrt {{{\left( 2 \right)}^4} - 12} \cr & m = 32 + 8 \cr & m = 40 \cr & {\text{Evaluate the function at }}x = 2 \cr & f\left( 2 \right) = {\left( 2 \right)^2}\sqrt {{{\left( 2 \right)}^4} - 12} \cr & f\left( 2 \right) = 8 \cr & {\text{we know the point }}\left( {2,8} \right){\text{ and the slope }}m = 40 \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 - 8 = 40\left( {x - 2} \right) \cr & {\text{simplifying}} \cr & y - 8 = 40x - 80 \cr & y = 40x - 72 \cr}