Derivative of x/(1+e^x)

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = x$ and $g{\left(x \right)} = e^{x} + 1$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

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

   To find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $e^{x} + 1$ term by term:

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

      2. The derivative of $e^{x}$ is itself.

      The result is: $e^{x}$

   Now plug in to the quotient rule:

   $\frac{- x e^{x} + e^{x} + 1}{\left(e^{x} + 1\right)^{2}}$

The answer is:

$$\frac{- x e^{x} + e^{x} + 1}{\left(e^{x} + 1\right)^{2}}$$