Derivative of log(x+1)/(x+1)

1. Apply the quotient rule, which is:

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

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

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

   1. Let $u = x + 1$.

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

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

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

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

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

         The result is: $1$

      The result of the chain rule is:

      $$\frac{1}{x + 1}$$

   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:

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

2. Now simplify:

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

The answer is: