You can differentiate this function by using the quotient rule, which tells you that you can differentiate functions that take the form
#f(x) = g(x)/(h(x))#, with #g(x)!=0#
by using the formula
#color(blue)(d/dx(f(x)) = ([d/dx(g(x))] * h(x) - g(x) * d/dx(h(x)))/[(g(x)]^2)#
In your case, you have #g(x) = e^x# and #h(x) = e^x + 1#, which means that you can write
#d/dx(f(x)) = ([d/dx(e^x)] * (e^x - 1) - e^x * d/dx(e^x - 1))/(e^x + 1)^2#
#f^' = (e^x * (e^x + 1) - e^(2x))/(e^x + 1)^2#
#f^' = (color(red)(cancel(color(black)(e^(2x)))) + e^x - color(red)(cancel(color(black)(e^(2x)))))/(e^x + 1)^2 = color(green)(e^x/(e^x + 1)^2)#
Alternatively, you can play around with the function a bit and write it as
#f(x) = e^x * (e^x + 1)^(-1)#
In this case, you can the product rule and the chain rule to differentiate the function
#d/dx(f(x)) = d/dx(e^x) * (e^x + 1)^(-1) + e^x * d/dx(e^x + 1)^(-1)#
#f^' = e^x * (e^x + 1)^(-1) + e^x * (-1) * (e^x + 1)^(-2) * e^x#
#f^' = e^x/(e^x + 1) - e^(2x)/(e^x + 1)^2#
Once again, you have
#f^' = (e^x * (e^x + 1) - e^(2x))/(e^x + 1)^2#
#f^' = (color(red)(cancel(color(black)(e^(2x)))) + e^x - color(red)(cancel(color(black)(e^(2x)))))/(e^x + 1)^2 = color(green)(e^x/(e^x + 1)^2)#
