Derivative of ze^z

1. Apply the product rule:

   $$\frac{d}{d z} f{\left(z \right)} g{\left(z \right)} = f{\left(z \right)} \frac{d}{d z} g{\left(z \right)} + g{\left(z \right)} \frac{d}{d z} f{\left(z \right)}$$

   $f{\left(z \right)} = z$; to find $\frac{d}{d z} f{\left(z \right)}$:

   1. Apply the power rule: $z$ goes to $1$

   $g{\left(z \right)} = e^{z}$; to find $\frac{d}{d z} g{\left(z \right)}$:

   1. The derivative of $e^{z}$ is itself.

   The result is: $z e^{z} + e^{z}$

2. Now simplify:

   $$\left(z + 1\right) e^{z}$$

The answer is:

$$\left(z + 1\right) e^{z}$$