How do you find the derivative of #[e^x / (1 - e^x)]#?

Question

18168 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-07T06:49:15

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

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

To find #y'#, we will use the Quotient Rule , which states, :

#(u/v)'=(vu'-uv')/v^2#.

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

#rArr y'={(1-e^x)(e^x)'-e^x(1-e^x)'}/(1-e^x)^2#.

#={(1-e^x)(e^x)-e^x(1'-(e^x)')}/(1-e^x)^2#.

#={(1-e^x)e^x-e^x(0-e^x)}/(1-e^x)^2#.

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

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

Enjoy Maths.!