Derivative of (1+x)ln(1+x)

1. Apply the product rule:

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

   $f{\left(x \right)} = x + 1$; to find $\frac{d}{d x} f{\left(x \right)}$:

   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: $1$

   $g{\left(x \right)} = \log{\left(x + 1 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

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

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

         The result is: $1$

      The result of the chain rule is:

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

   The result is: $\log{\left(x + 1 \right)} + 1$

The answer is:

$$\log{\left(x + 1 \right)} + 1$$