Derivative of x+5+6ln((x)/(x-1))

1. Differentiate $\left(x + 5\right) + 6 \log{\left(\frac{x}{x - 1} \right)}$ term by term:

   1. Differentiate $x + 5$ term by term:

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

      2. The derivative of the constant $5$ is zero.

      The result is: $1$

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

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

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

      3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \frac{x}{x - 1}$:

         1. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = x$ and $g{\left(x \right)} = x - 1$.

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

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

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

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

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

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

               The result is: $1$

            Now plug in to the quotient rule:

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

         The result of the chain rule is:

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

      So, the result is: $- \frac{6 \left(x - 1\right)}{x \left(x - 1\right)^{2}}$

   The result is: $1 - \frac{6 \left(x - 1\right)}{x \left(x - 1\right)^{2}}$

2. Now simplify:

   $$\frac{x \left(x - 1\right) - 6}{x \left(x - 1\right)}$$

The answer is:

$$\frac{x \left(x - 1\right) - 6}{x \left(x - 1\right)}$$