Derivative of cx^2*exp^(-3/x)

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = c x^{2}$ and $g{\left(x \right)} = e^{\frac{3}{x}}$.

   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: $x^{2}$ goes to $2 x$

      So, the result is: $2 c x$

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

   1. Let $u = \frac{3}{x}$.

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

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

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

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

         So, the result is: $- \frac{3}{x^{2}}$

      The result of the chain rule is:

      $$- \frac{3 e^{\frac{3}{x}}}{x^{2}}$$

   Now plug in to the quotient rule:

   $\left(2 c x e^{\frac{3}{x}} + 3 c e^{\frac{3}{x}}\right) e^{- \frac{6}{x}}$

2. Now simplify:

   $$c \left(2 x + 3\right) e^{- \frac{3}{x}}$$

The answer is:

$$c \left(2 x + 3\right) e^{- \frac{3}{x}}$$