Question

Find derivative of the function:
y=ln(e^x/(e^x-1))

Answer 1

using the chain rule, since y = ln(u)
y' = 1/u u'
Now, since u = v/w,
u' = (v'w - vw')/w^2
= (e^x(e^x-1) - e^x(e^x))/(e^x-1)^2
= (e^2x - e^x - e^2x)/(e^x-1)^2
= -e^x/(e^x-1)^2
So,
y' = (e^x-1)/e^x * -e^x/(e^x-1)^2
= -1/(e^x-1)

Or, since y = ln(e^x) - ln(e^x-1) = x - ln(e^x-1)
y' = 1 - e^x/(e^x-1) = -1/(e^x-1)