Derivative of (x-1)e^x

1. Apply the product rule:

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

   $f{\left(x \right)} = x - 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $x - 1$ term by term:

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

      2. The derivative of the constant $-1$ is zero.

      The result is: $1$

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

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

   The result is: $e^{x} + \left(x - 1\right) e^{x}$

2. Now simplify:

   $$x e^{x}$$

The answer is: