Derivative of 2exp(x-2)*ln(x-2)

1. Apply the product rule:

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

   $f{\left(x \right)} = 2 e^{x - 2}$; 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. Let $u = x - 2$.

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

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

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

            1. Apply the power rule: $x$ goes to $1$

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

            The result is: $1$

         The result of the chain rule is:

         $$e^{x - 2}$$

      So, the result is: $2 e^{x - 2}$

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

   1. Let $u = x - 2$.

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

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

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

         1. Apply the power rule: $x$ goes to $1$

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

         The result is: $1$

      The result of the chain rule is:

      $$\frac{1}{x - 2}$$

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

2. Now simplify:

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

The answer is: