質問: How do you find the derivative of #[e^x / (1 - e^x)]#? ======================================================
回答: #y'=e^x/(1-e^x)^2#. #### Explanation: 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.!