How do you find the derivative of e^x / 1+ e^x?
Thanks for the help!
How do you find the derivative of e^x / 1+ e^x?
2011-06-17 11:43 am
回答 (5)
2011-06-17 11:47 am
✔ 最佳答案
Use quotient rule of differentiation,e^x / (1 + e^x)
[ e^x(1 + e^x) - e^x(e^x) ] / (1 + e^x)^2
e^x / (1 + e^x)^2
2011-06-17 12:08 pm
Apply division rule of derivative.
d/dx [ u*v ] = {v*d/dx [ u ] - u*d/dx [ v ]} / v^2
Hence d/dx [ e^x / 1+ e^x ] = {[1+ e^x ] * d/dx[ e^x] - e^x * d/dx[1+e^x ]} / [1+e^x]^2
Simplifying this we get an answer as ---- e^x / [1+e^x]^2
d/dx [ u*v ] = {v*d/dx [ u ] - u*d/dx [ v ]} / v^2
Hence d/dx [ e^x / 1+ e^x ] = {[1+ e^x ] * d/dx[ e^x] - e^x * d/dx[1+e^x ]} / [1+e^x]^2
Simplifying this we get an answer as ---- e^x / [1+e^x]^2
參考: I.L.Maths
2011-06-17 12:11 pm
( 1 + e^x) (e^x) - e^x(e^x) / (1+e^x)^2
= e^x / (1+e^x)^2
= e^x / (1+e^x)^2
2011-06-17 12:06 pm
f (x) = e^x / ( 1 + e^x )
f ` (x) = ln ( 1 + e^x )
f ` (x) = ln ( 1 + e^x )
2011-06-17 11:48 am
How do you find the derivative of e^x / 1+ e^x?
Sol put (1+e^x) = u
find derivative
(0+e^x)dx = du
integrate[du/u]
log u + C
log(1+e^x) + C ......................Ans
Sol put (1+e^x) = u
find derivative
(0+e^x)dx = du
integrate[du/u]
log u + C
log(1+e^x) + C ......................Ans
收錄日期: 2021-05-01 00:53:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110617034330AAKgGFm