Derivative of 2*x*e^(-x^2)

1. Apply the quotient rule, which is:

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

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

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

   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: $2$

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

   1. Let $u = x^{2}$.

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

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

      1. Apply the power rule: $x^{2}$ goes to $2 x$

      The result of the chain rule is:

      $$2 x e^{x^{2}}$$

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is: