Derivative of ln(sqrt(2x/x+1))

1. Let $u = \sqrt{1 + \frac{2 x}{x}}$.

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

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sqrt{1 + \frac{2 x}{x}}$:

   1. Let $u = 1 + \frac{2 x}{x}$.

   2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(1 + \frac{2 x}{x}\right)$:

      1. Differentiate $1 + \frac{2 x}{x}$ term by term:

         1. Apply the quotient rule, which is:

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

            $f{\left(x \right)} = 2 x$ and $g{\left(x \right)} = x$.

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

               So, the result is: $2$

            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:

            $0$

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

         The result is: $0$

      The result of the chain rule is:

      $$0$$

   The result of the chain rule is:

   $$0$$

The answer is:

$$0$$