## Calculus with Applications (10th Edition)

$p^{\prime}(z)=(6z+1)^{1/3}(14z+1)$
$p(z)=z\cdot(6z+1)^{4/3}=u(z)\cdot v(z)$ Use the product rule $\color{blue}{[u(z)\cdot v(z)]^{\prime}=u(z)\cdot v^{\prime}(z)+v(z)\cdot u^{\prime}(z)}$ For $v^{\prime}(z)$, we need the chain rule. $\left[\begin{array}{llll} u(z)=z & , & v(z)=(6z+1)^{4/3} & \\ u^{\prime}(z)=1 & & v(z)=[w(z)]^{4/3} & \\ & & v^{\prime}(z)=\frac{4}{3}(6z+1)^{1/3}\cdot w^{\prime}(z) & (\text{chain rule)})\\ & & v^{\prime}(z)=\frac{4}{3}(6z+1)^{1/3}\cdot 6 & \end{array}\right]$. $p^{\prime}(z)=u(z)\cdot v^{\prime}(z)+v(z)\cdot u^{\prime}(z)$ $=z\displaystyle \cdot\frac{4}{3}(6z+1)^{1/3}\cdot 6+(6z+1)^{4/3}\cdot 1$ $=8z(6z+1)^{1/3}+(6z+1)^{4/3}$ $=(6z+1)^{1/3}[8z+(6z+1)^{3/3}]$ $=(6z+1)^{1/3}[8z+6z+1]$ $=(6z+1)^{1/3}(14z+1)$