Derivative of xe^(2x-1)

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 $1$

   $g{\left(x \right)} = e^{2 x - 1}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x - 1$.

   2. The derivative of $e^{u}$ is itself.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x - 1\right)$:

      1. Differentiate $2 x - 1$ term by term:

         1. The derivative of a constant times a function is the constant times the derivative of the function.

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

            So, the result is: $2$

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

         The result is: $2$

      The result of the chain rule is:

      $$2 e^{2 x - 1}$$

   The result is: $2 x e^{2 x - 1} + e^{2 x - 1}$

2. Now simplify:

   $$\left(2 x + 1\right) e^{2 x - 1}$$

The answer is:

$$\left(2 x + 1\right) e^{2 x - 1}$$