Question

find y'. y=ln e^x/e^x-1

Answer 1

is this written correctly:

\[y = \ln \frac{ e^{x} }{ e^{x} - 1 }\]

Answer 2

yes it is

Answer 3

\[\ln u = \frac{ du }{ u }\]

this is a simplified version of it

use the chain rule here where u = e^x / (e^x - 1) and du = the derivative of that

Answer 4

I tried that but got stuck trying to work out the derivative of e^x and e^x-1

Answer 5

\[\frac{ d }{ dx } \frac{ e^{x} }{ e^{x} - 1 }\]

i was taught in high school "down d up - up d down all over down squared"

just means "denominator * derivative of numerator - numerator * derivative of denominator, all divided by the square of the denominator"

in this case, the derivative of the numerator is e^x, and the derivative of the denominator is also e^x so we have

\[\frac{ d }{ dx }\frac{ e^{x} }{ e^{x} - 1 } = \frac{ (e^{x} - 1) * e^{x} - e^{x} * e^{x} }{ (e^{x} - 1)^{2} }\]

Answer 6

that simplifies to:

Answer 7

\[-\frac{ e^{x} }{ (e^{x} - 1)^{2} }\]

Answer 8

do you mind breaking down how you go t that answer. please :)

Answer 9

that's just the derivative of e^x / (e^x - 1)
you use the quotient rule for derivatives

Answer 10

and we said u = e^x / (e^x - 1), and du, we just solved for, du = -e^x / (e^x - 1)^2

therefore the derivative of ln u = du/u =

\[\frac{ -e^{x} (e^{x} - 1) }{ e^{x}(e^{x} - 1)^{2} } = \frac{ -1 }{ e^{x} - 1 } = \frac{ 1 }{ 1 - e^{x} }\]

Answer 11

okay i think i got it . thank you so much