Derivative of ln((x+9)/x)-1

1. Differentiate $\log{\left(\frac{x + 9}{x} \right)} - 1$ term by term:

   1. Let $u = \frac{x + 9}{x}$.

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \frac{x + 9}{x}$:

      1. Apply the quotient rule, which is:

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

         $f{\left(x \right)} = x + 9$ and $g{\left(x \right)} = x$.

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

         1. Differentiate $x + 9$ term by term:

            1. The derivative of the constant $9$ is zero.

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

            The result is: $1$

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

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

         Now plug in to the quotient rule:

         $- \frac{9}{x^{2}}$

      The result of the chain rule is:

      $$- \frac{9}{x \left(x + 9\right)}$$

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

   The result is: $- \frac{9}{x \left(x + 9\right)}$

2. Now simplify:

   $$- \frac{9}{x \left(x + 9\right)}$$

The answer is:

$$- \frac{9}{x \left(x + 9\right)}$$