Derivative of te^t

1. Apply the product rule:

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

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

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

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

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

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

2. Now simplify:

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

The answer is:

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