高等数学隐函数题高等数学题.跪求第一大题所有小题的过程.悬赏酌情考虑.

学习 时间:2026-04-04 18:23:07 阅读:7172
高等数学隐函数题高等数学题.跪求第一大题所有小题的过程.悬赏酌情考虑.

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殷勤的草莓

寂寞的网络

2026-04-04 18:23:07

每道题都是对等式两边的x求导
(1)2x-2yy'+2y+2xy'=2
(x-y)y'=1-x-y
y'=(1-x-y)/(x-y)
(2)y'=1+y'/y
yy'=y+y'
y'=y/(y-1)
(3)e^(x+y)*(1+y')=y+xy'
[e^(x+y)-x]y'=y-e^(x+y)
y'=[y-e^(x+y)]/[e^(x+y)-x]
(4)[(xy'-y)/x^2]/(1+y^2/x^2)=[(2x+2yy')/2√(x^2+y^2)]/√(x^2+y^2)
(xy'-y)/(x^2+y^2)=(x+yy')/(x^2+y^2)
xy'-y=x+yy'
y'=(x+y)/(x-y)
(5)e^(xy)*(y+xy')+y'lnx+y/x=-2sin2x
[e^(xy)*x+lnx]y'=-2sin2x-e^(xy)*y-y/x
y'=-[2sin2x+e^(xy)*y+y/x]/[e^(xy)*x+lnx]
(6)e^y+e^y*y'x+2yy'=0
(e^y*x+2y)y'=-e^y
y'=-(e^y)/(e^y*x+2y)
(7)1-cos(y/x)*(y'x-y)/x^2=0
y'x-y=x^2/cos(y/x)
y'=x*sec(y/x)+y/x
(8)-sin(x^2-y)*(2x-y')=1
y'-2x=csc(x^2-y)
y'=2x+csc(x^2-y)
再问: 我记得第一题的答案好像是分数。。
再答: 客气

最新回答共有2条回答

  • 矮小的天空
    回复
    2026-04-04 18:23:07

    每道题都是对等式两边的x求导(1)2x-2yy'+2y+2xy'=2(x-y)y'=1-x-yy'=(1-x-y)/(x-y)(2)y'=1+y'/yyy'=y+y'y'=y/(y-1)(3)e^(x+y)*(1+y')=y+xy'[e^(x+y)-x]y'=y-e^(x+y)y'=[y-e^(x+y)]/[e^(x+y)-x](4)[(xy'-y)/x^2]/(1+y^2/x^2)=[(2x+2yy')/2√(x^2+y^2)]/√(x^2+y^2)(xy'-y)/(x^2+y^2)=(x+yy')/(x^2+y^2)xy'-y=x+yy'y'=(x+y)/(x-y)(5)e^(xy)*(y+xy')+y'lnx+y/x=-2sin2x[e^(xy)*x+lnx]y'=-2sin2x-e^(xy)*y-y/xy'=-[2sin2x+e^(xy)*y+y/x]/[e^(xy)*x+lnx](6)e^y+e^y*y'x+2yy'=0(e^y*x+2y)y'=-e^yy'=-(e^y)/(e^y*x+2y)(7)1-cos(y/x)*(y'x-y)/x^2=0y'x-y=x^2/cos(y/x)y'=x*sec(y/x)+y/x(8)-sin(x^2-y)*(2x-y')=1y'-2x=csc(x^2-y)y'=2x+csc(x^2-y) 再问: 我记得第一题的答案好像是分数。。 再答: 客气

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