Derivative of x*ln(1-x)

1. Apply the product rule:

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

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

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

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

   1. Let $u = 1 - x$.

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

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

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

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

         2. 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: $-1$

         The result is: $-1$

      The result of the chain rule is:

      $$- \frac{1}{1 - x}$$

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

2. Now simplify:

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

The answer is:

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