Question

求第3，4题的答案

Answer 1

(3)
y^2 + sinx + e^(x-y) =1
2y.y' + cosx + (1-y') e^(x-y) =0
[2y- e^(x-y) ]y' = -[ cosx + e^(x-y) ]
y' = -[ cosx + e^(x-y) ]/ [2y- e^(x-y) ]
(4)
e^(x+y) -y +x^2 =1
(1+y' ).e^(x+y) - y' +2x =0
[e^(x+y) -1]y' = -[ e^(x+y) + 2x ]

y' = -[ e^(x+y) + 2x ]/[e^(x+y) -1]

Answer 2

3、方程两边对x求导
2yy'+cosx+e^(x-y)*(1-y')=0
[2y-e^(x-y)]y'=-cosx-e^(x-y)
y'=[cosx+e^(x-y)]/[e^(x-y)-2y]
4、方程两边对x求导
e^(x+y)*(1+y')-y'+2x=0
[e^(x+y)-1]y'=-2x-e^(x+y)
y'=[2x+e^(x+y)]/[1-e^(x+y)]