Derivative of x*3^(4-2x)

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 $1$

   $g{\left(x \right)} = 3^{4 - 2 x}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 4 - 2 x$.

   2. $\frac{d}{d u} 3^{u} = 3^{u} \log{\left(3 \right)}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(4 - 2 x\right)$:

      1. Differentiate $4 - 2 x$ term by term:

         1. The derivative of the constant $4$ 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: $-2$

         The result is: $-2$

      The result of the chain rule is:

      $$- 2 \cdot 3^{4 - 2 x} \log{\left(3 \right)}$$

   The result is: $- 2 \cdot 3^{4 - 2 x} x \log{\left(3 \right)} + 3^{4 - 2 x}$

2. Now simplify:

   $$81 \cdot 3^{- 2 x} \left(- 2 x \log{\left(3 \right)} + 1\right)$$

The answer is:

$$81 \cdot 3^{- 2 x} \left(- 2 x \log{\left(3 \right)} + 1\right)$$