How do you differentiate  f(x)=ln (e^x/(e^x+1))?

May 8, 2017

$f ' \left(x\right) = \frac{1}{{e}^{x} + 1}$

Explanation:

This can be rewritten as

$f \left(x\right) = \ln \left({e}^{x}\right) - \ln \left({e}^{x} + 1\right)$

$f \left(x\right) = x \ln e - \ln \left({e}^{x} + 1\right)$

We know that $\ln \left(e\right) = 1$.

$f \left(x\right) = x - \ln \left({e}^{x} + 1\right)$

Now use the chain and power rule to differentiate.

$f ' \left(x\right) = 1 - \frac{{e}^{x}}{{e}^{x} + 1}$

$f ' \left(x\right) = \frac{{e}^{x} + 1 - {e}^{x}}{{e}^{x} + 1}$

$f ' \left(x\right) = \frac{1}{{e}^{x} + 1}$

Hopefully this helps!