Thomas' Calculus 13th Edition

First Derivative: $y'=x^{2}+x+\frac{1}{4}$ Second Derivative: $y''=2x+1$
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)