Derivative of xe^x-1

1. Differentiate $x e^{x} - 1$ term by term:

   1. Apply the product rule:

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

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

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

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

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

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

   2. The derivative of the constant $\left(-1\right) 1$ is zero.

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

2. Now simplify:

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

The answer is: