## Calculus (3rd Edition)

(a) $\frac{d}{dt}ct^3=3ct^2$. (b) $\frac{d}{dz}(5z+4cz^2)=5+8cz$. (c) $\frac{d}{dy}(9c^2y^3-24c)=27c^3y^2$.
Since $c$ is constant, we have (a) $\frac{d}{dt}ct^3=3ct^2$. (b) $\frac{d}{dz}(5z+4cz^2)=5+8cz$. (c) $\frac{d}{dy}(9c^2y^3-24c)=27c^3y^2$.