Derivative of 5xe^(-2x)

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

   1. Apply the quotient rule, which is:

      $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

      $f{\left(x \right)} = x$ and $g{\left(x \right)} = e^{2 x}$.

      To find $\frac{d}{d x} f{\left(x \right)}$:

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

      To find $\frac{d}{d x} g{\left(x \right)}$:

      1. Let $u = 2 x$.

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$:

         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: $2$

         The result of the chain rule is:

         $$2 e^{2 x}$$

      Now plug in to the quotient rule:

      $\left(- 2 x e^{2 x} + e^{2 x}\right) e^{- 4 x}$

   So, the result is: $5 \left(- 2 x e^{2 x} + e^{2 x}\right) e^{- 4 x}$

2. Now simplify:

   $$5 \cdot \left(1 - 2 x\right) e^{- 2 x}$$

The answer is:

$$5 \cdot \left(1 - 2 x\right) e^{- 2 x}$$