Answer
$$\frac{{dy}}{{dx}} = 2x - \frac{7}{{{x^2}}}{\text{ and }}\frac{{{d^2}y}}{{d{x^2}}} = 2 + \frac{{14}}{{{x^3}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^3} + 7}}{x} \cr
& {\text{Use the distributive property}} \cr
& y = \frac{{{x^3}}}{x} + \frac{7}{x} \cr
& {\text{where }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& y = {x^2} + 7{x^{ - 1}} \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2} + 7{x^{ - 1}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& \frac{{dy}}{{dx}} = 2x + 7\left( { - {x^{ - 2}}} \right) \cr
& \frac{{dy}}{{dx}} = 2x - 7{x^{ - 2}} \cr
& \frac{{dy}}{{dx}} = 2x - \frac{7}{{{x^2}}} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {2x - 7{x^{ - 2}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 2\left( 1 \right) - 7\left( { - 2{x^{ - 3}}} \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 2 + 14{x^{ - 3}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 2 + \frac{{14}}{{{x^3}}} \cr
& \cr
& \frac{{dy}}{{dx}} = 2x - \frac{7}{{{x^2}}}{\text{ and }}\frac{{{d^2}y}}{{d{x^2}}} = 2 + \frac{{14}}{{{x^3}}} \cr} $$
