# How do you find the first derivative of f(z) = z^2 e^(2z)?

Apr 12, 2018

$f ' \left(z\right) = 2 z {e}^{2 z} \left(z + 1\right)$

#### Explanation:

The Product Rule tells us that if we have a multiplication of functions $\left(f g\right) ,$ then

$\left(f g\right) ' = f g ' + g f '$

Here,

$f \left(z\right) = {z}^{2} {e}^{2 z}$

$f ' \left(z\right) = {z}^{2} \frac{d}{\mathrm{dz}} {e}^{2 z} + {e}^{2 z} \frac{d}{\mathrm{dz}} {z}^{2}$

$\frac{d}{\mathrm{dz}} {z}^{2} = \mathrm{dz}$

$\frac{d}{\mathrm{dz}} {e}^{2 z} = 2 {e}^{2 z}$

As per the Product Rule and Chain Rule. So,

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

We may simplify a bit.

$f ' \left(z\right) = 2 z {e}^{2 z} \left(z + 1\right)$