Answer
First Derivative: $y'=x^{2}+x+\frac{1}{4}$
Second Derivative: $y''=2x+1$
Work Step by Step
Using the power rule:
First Derivative:
$y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{4}$
$y'=\frac{(3)x^{3-1}}{3}+\frac{(2)x^{2-1}}{2}+\frac{(1)x^{1-1}}{4}$
$y'=x^{2}+x+\frac{1}{4}$
Second Derivative:
$y'=x^{2}+x+\frac{1}{4}$
$y''=(2)x^{2-1}+(1)x^{1-1}+(0)$
$y''=2x+1$
(the derivative of $\frac{1}{4}$ will be 0 because of the rule of the constant function)
