高等数学隐函数题

Question

高等数学题。。。跪求第一大题所有小题的过程。。。。悬赏酌情考虑。。

Answer 1

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)追问

我记得第一题的答案好像是分数。。

追答

客气

Answer 2

(1) 2x - 2y*y' + 2y + 2xy' = 2
y'= (1 - x - y)/( x - y )
(2) y' = 1 + 1/y * y'
y' = 1/(1-1/y)
(3) e^(x+y) * (1+y') = y + xy'
y' = [y - e^(x+y)] / [e^(x+y) - x]
(4)1/[1+(y/x)^2] * (y'/x - y/x^2) = 1/[2*(x^2+y^2)] * (2x + 2y*y')
(y'x-y) = x + y*y'
y' = (x+y)/(x-y)
(5) e^xy * (y+xy') + y/x + y' * lnx = -2sin2x
y' = (-2sin2x - y* e^xy - y/x)/(x* e^xy + lnx)
(6) e^y * y' *x + e^y + 2y*y' = 0
y' = -e^y / (x*e^y + 2y)
(7) 1 - cos(y/x) * (y'x-y)/x^2 =0
y'=x / cos(y/x) + y/x
(8) -sin(x^2 -y) * (2x-y') = 1
y' = 1 / sin(x^2 - y) + 2x

Answer 3

孩纸，这也太多了吧。。。。

而且到底是X对Y求导，还是Y对X求导，还是两个都要，好多

Answer 4

两边对x求导，再整理一下就得出答案了呀，看看例题不就会了吧，导数是高数里的基础。比如，第二题两边对x求导后，整理得y'=y/y-1.