In taking a derivative, I end up with this epression:
$e^x*(1-e^x)-(1+e^x)*(-e^x) / (1-e^x)^2$
However, from a calculator I see that the first line simplifies in nicer, simpler expression:
$1/cosh(x)-1$
I don't have yet a knowledge of hyperbolic functions. Could you show me and explain me this simplification, please?
I assume you are going from $$ g(x) = \dfrac{1+\mathrm{e}^x}{1-\mathrm{e}^x} $$ and taking the derivative?
in any case $$ \dfrac{\mathrm{e}^x}{1-\mathrm{e}^x}+\dfrac{\mathrm{e}^x(1+\mathrm{e}^x)}{(1-\mathrm{e}^x)^2}=\dfrac{\mathrm{e}^x(1-\mathrm{e}^x) +\mathrm{e}^x(1+\mathrm{e}^x)}{(1-\mathrm{e}^x)^2}= \dfrac{2\mathrm{e}^x}{(1-\mathrm{e}^x)^2} $$ this equivalent to $$ \dfrac{2\mathrm{e}^x}{1-2\mathrm{e}^x+\mathrm{e}^{2x}}= \frac{2\mathrm{e}^{x}}{2\mathrm{e}^{x}\left(\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}-1\right)}=\dfrac{1}{\cosh(x)-1} $$