Derivative of 6*x*ln(1+x)

1. Apply the product rule:

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

   $f{\left(x \right)} = 6 x$; 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: $6$

   $g{\left(x \right)} = \log{\left(x + 1 \right)}$; 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: $\frac{6 x}{x + 1} + 6 \log{\left(x + 1 \right)}$

2. Now simplify:

   $$\frac{6 \left(x + \left(x + 1\right) \log{\left(x + 1 \right)}\right)}{x + 1}$$

The answer is:

$$\frac{6 \left(x + \left(x + 1\right) \log{\left(x + 1 \right)}\right)}{x + 1}$$