这些函数的微分怎么求啊？

Question

如题所述

Answer 1

这些函数的的微分怎么求？两种方法①复合函数的微分，可以象求导一样，从外函数到内函数一层一层地往里求，类似于求导的链式法则；②先求函数的导数，再求函数的微分。追答

Answer 2

求下列函数的微分：
(1).y=ln(1+e^x)
dy/dx=1/(1+e^x)*e^x
dy=〔e^x/(1+e^x)〕dx
(2).y=ln(x+1)+1/x-2√x
dy/dx=1/(x+1)-1/x^2-2*1/2√x
=1/(x+1)-1/x^2-1/√x
dy=〔1/(x+1)-1/x^2-1/√x〕dx
(3).y=e^(2x)/x
dy/dx=〔2e^(2x)*x-e^(2x)*1〕/x^2
=〔e^(2x)(2x-1)〕/x^2
dy={〔e^(2x)(2x-1)〕/x^2}dx
(4).y=x/√(x^2+1)
dy/dx=〔√(x^2+1)-x*2x/2√(x^2+1)〕/(x^2+1)
=〔(x^2+1)-x^2〕/〔(x^+1)√(x^2+1)〕
=1/〔(x^2+1)√(x^2+1)〕
dy=dx/〔(x^+1)√(x^2+1)〕

Answer 3

y = f(x), dy = f'(x)dx
(1) dy = [e^x/(1+e^x)]dx
(2) dy = [1/(x+1)-1/x^2-1/x^(3/2)]dx
(3) dy = [(2x-1)e^(2x)/x^2]dx
(4) dy = [1/(x^2+1)^(3/2)]dx

Answer 4

(1)
y=ln(1+e^x)
dy
=[1/(1+e^x)] d(1+e^x)
=[1/(1+e^x)] e^x dx
=[e^x/(1+e^x)] dx
(2)
y=ln(x+1)+1/x-2√x
dy
=d[ln(x+1)+1/x-2√x]
= (1/(x+1) - 1/x^2 - 1/√x) dx
(3)
y=e^(2x)/x
dy
=[xde^(2x) - e^(2x) dx ]/x^2
=[xe^(2x) d(2x) - e^(2x) dx ]/x^2
=[xe^(2x) (2dx) - e^(2x) dx ]/x^2
=[e^(2x) .(2x -1)/x^2] dx
(4)

y=x/√(x^2+1)
dy
=[√(x^2+1) dx - xd√(x^2+1) ]/(x^2+1)
=[√(x^2+1) dx - x [x/√(x^2+1)] dx ] /(x^2+1)
=[(x^2+1) dx - x^2 dx ] /(x^2+1)^(3/2)
= dx/(x^2+1)^(3/2)本回答被提问者采纳