How do you find the derivative of #y=ln((1+e^x)/(1-e^x))#?
1 Answer
Dec 17, 2016
Explanation:
Let
Then
#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!