How do you find the derivative of #y=ln(1+e^x/1-e^x)#?

2 Answers

I think that this was not the function that you intended to write (probably you omitted parentheses, which are essential when one writes in-line fractions). In fact, you can rewrite that function as follows:
#c(x)=ln(1+e^x/1-e^x)=ln(1+e^x-e^x)=ln(1)=0#
The derivative of a constant function is #0#, so
#c'(x)=0#


I'll try to infer the function that you wanted to write:
#f(x)=ln((1+e^x)/(1-e^x))#
The code to render this properly is f(x)=ln((1+e^x)/(1-e^x)) (notice the usage of parentheses).

This is a function composition. Given two functions #y=g(x)# and #z=h(y)# you can compute the derivative of the composition #f(x)=h(g(x))# as follows:
#f'(x)=h'(g(x))*g'(x)#
In this particular case #g(x)=(1+e^x)/(1-e^x)# and #h(y)=ln(y)#.

The derivative of #g# we is the derivative of a quotient
#g'(x)=(e^x(1-e^x)-(1+e^x)(-e^x))/(1-e^x)^2=(2e^x)/(1-e^x)^2#
and the derivative of #h# is the derivative of the logarithmic function
#h'(y)=1/y#
So
#f'(x)=h'(g(x)) * g'(x)=1/((1+e^x)/(1-e^x)) * (2e^x)/(1-e^x)^2=(2e^x)/((1+e^x)(1-e^x))=(2e^x)/(1-e^(2x))#

Apr 26, 2015

I assume you want to differentiate:

#y=ln((1+e^x)/(1-e^x))#.

You could use #d/dx (lnu) = 1/u (du)/dx# with #u = (1+e^x)/(1-e^x)# and use the quotient rule to find #(du)/dx#, but it's a little quicker to use #ln(a/b) = lna - lnb# to rewrite the function as:

#y=ln(1+e^x) - ln(1-e^x)#. before differentiating:

#y' = 1/(1+e^x) e^x - 1/(1-e^x) (-e^x)#

Or, simplifying:

#y' = e^x/(1+e^x) + e^x/(1-e^x) = (2e^x)/(1-e^(2x))#