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

1 Answer
Dec 17, 2016

#dy/dx= (2e^x)/((1 + e^x)(1 - e^x)) = (2e^x)/(1 - e^(2x))#

Explanation:

Let #y= lnu# and #u = (1 + e^x)/(1 - e^x)#.

Then #dy/(du) = 1/u# and #(du)/dx = (e^x(1 - e^x) - (-e^x(1 + e^x)))/(1 - e^x)^2#

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = 1/u xx (e^x - e^(2x) + e^x + e^(2x))/(1 - e^x)^2#

#dy/dx= 1/u xx (2e^x)/(1 - e^x)^2#

#dy/dx = (2e^x)/((1 + e^x)/(1 - e^x)) xx 1/(1 - e^x)^2#

#dy/dx = (2e^x(1 - e^x))/((1 + e^x)(1 - e^x)^2)#

#dy/dx= (2e^x)/((1 + e^x)(1 - e^x))#

Hopefully this helps!