How do you find the derivative of g(z)=(z^2+3)e^z?

Sep 9, 2017

$g ' \left(z\right) = \left({z}^{2} + 2 z + 3\right) {e}^{z}$

Explanation:

we will use the product rule

$g \left(z\right) = u \left(z\right) v \left(z\right) \implies g ' \left(z\right) = \textcolor{red}{u ' \left(z\right)} v \left(z\right) + u \left(z\right) \textcolor{b l u e}{v ' \left(z\right)}$

we have

$g \left(z\right) = \left({z}^{2} + 3\right) {e}^{z}$

$u \left(z\right) = {z}^{2} + 3 \implies \textcolor{red}{u ' \left(z\right) = 2 z}$

$v \left(z\right) = {e}^{z} \implies v ' \left(z\right) = \textcolor{b l u e}{{e}^{z}}$

$\therefore g ' \left(z\right) = \textcolor{red}{2 z} {e}^{z} + \left({z}^{2} + 3\right) \textcolor{b l u e}{{e}^{z}}$

simplifying and tidying up

$g ' \left(z\right) = {e}^{z} \left(2 z + {z}^{2} + 3\right)$

$g ' \left(z\right) = \left({z}^{2} + 2 z + 3\right) {e}^{z}$