$$f(z) = \frac{\exp(z)}{1+\exp(z)}$$

My thought was to apply quotient rule with the denominator needing chain rule. But I feel like my answer is off.

Here's my working:

$$\frac{exp(z)}{1+exp(z)} = \frac{(1 + exp(z))^{-1}exp(z) - exp(z)(-1)(1 + exp(z))^{-2}(exp(z))}{(1 + exp(z))^2}$$

$$= \frac{exp(z)}{(1+exp(z))^3} + \frac{exp(z)^2}{(1+exp(z))^4}$$ $$= \frac{exp(z) + exp(z)^2 + exp(z)^2}{(1 + exp(z))^4}$$ $$= \frac{exp(z)(1 + 2exp(z))}{(1 + exp(z))^4}$$