Derivative of e^(2x-4)+2lnx

1. Differentiate $e^{2 x - 4} + 2 \log{\left(x \right)}$ term by term:

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

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

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

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

         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: $2$

         2. The derivative of the constant $-4$ is zero.

         The result is: $2$

      The result of the chain rule is:

      $$2 e^{2 x - 4}$$

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

      1. The derivative of $\log{\left(x \right)}$ is $\frac{1}{x}$.

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

   The result is: $2 e^{2 x - 4} + \frac{2}{x}$

2. Now simplify:

   $$\frac{2 \left(x e^{2 x - 4} + 1\right)}{x}$$

The answer is:

$$\frac{2 \left(x e^{2 x - 4} + 1\right)}{x}$$