Derivative of (3-x)*e^(2-x)

1. Apply the product rule:

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

   $f{\left(x \right)} = 3 - x$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $3 - x$ term by term:

      1. The derivative of the constant $3$ is zero.

      2. 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: $-1$

      The result is: $-1$

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

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

         1. The derivative of the constant $2$ is zero.

         2. 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: $-1$

         The result is: $-1$

      The result of the chain rule is:

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

   The result is: $- \left(3 - x\right) e^{2 - x} - e^{2 - x}$

2. Now simplify:

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

The answer is:

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