Answer
$y'=\frac{2x^3-7}{x^2}$
$y''=\frac{2x^3+14}{x^3}$
Work Step by Step
$$y=\frac{x^3+7}{x}$$
1) First derivative:
Apply the Derivative Quotient Rule: $$y'=\frac{x(x^3+7)'-(x)'(x^3+7)}{x^2}=\frac{x(3x^2+0)-1\times(x^3+7)}{x^2}$$
$$y'=\frac{3x^3-x^3-7}{x^2}=\frac{2x^3-7}{x^2}$$
2) Second derivative:
Apply the Derivative Quotient Rule:
$$y''=\frac{x^2(2x^3-7)'-(x^2)'(2x^3-7)}{x^4}=\frac{x^2(6x^2-0)-2x(2x^3-7)}{x^4}$$
$$y''=\frac{6x^4-4x^4+14x}{x^4}=\frac{2x^4+14x}{x^4}=\frac{2x^3+14}{x^3}$$
