Derivative of log(x+1)+1

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

   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:

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

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

   The result is: $\frac{1}{x + 1}$

2. Now simplify:

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

The answer is:

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