Derivative of y=ln(1+x)/1-x

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

   1. The derivative of a constant times a function is the constant times the derivative of the function.

      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:

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

      So, the result is: $\frac{1}{x + 1}$

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

2. Now simplify:

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

The answer is:

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