## Calculus (3rd Edition)

$$y'=15 x^4-14x.$$
Recall that $(x^n)'=nx^{n-1}$ Since $y=3x^5-7x^2+4$, then the derivative $y'$ is given by $$y'=3\times5x^{5-1}-2\times7x^{2-1}+0$$ $$=15 x^4-14x$$