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

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Answer 1 · 2017-05-08T13:35:16

$f'(x) = 1/(e^x + 1)$

This can be rewritten as

$f(x) = ln(e^x) - ln(e^x + 1)$

$f(x) = xlne - ln(e^x + 1)$

We know that $ln(e) = 1$ .

$f(x) = x - ln(e^x + 1)$

Now use the chain and power rule to differentiate.

$f'(x) = 1 - (e^x)/(e^x+ 1)$

$f'(x) = (e^x + 1 - e^x)/(e^x + 1)$

$f'(x) = 1/(e^x + 1)$

Hopefully this helps!

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