Question

隐函数求导

老哥们，我会把他展开，但是整理y导数的形式就不会了，麻烦过程详细点

Answer 1

全微分或者隐函数的导数公式，下面用全微分法，
d[cos(2y)+ln(x+y)-2xe^y]
=-2sin(2y)dy+(dx+dy)/(x+y)-2e^ydx-2xe^ydy=0
[1/(x+y)-2e^y]dx=[2sin(2y)-1/(x+y)+2xe^y]dy
y'(x)=dy/dx=[1/(x+y)-2e^y]/[2sin(2y)-1/(x+y)+2xe^y]
=[1-2(x+y)e^y]/[2(x+y)sin(2y)-1+2x(x+y)e^y]

Answer 2

cos2y+ln(x+y)-2xe^y=0

等式方程对x求导
-2sin2y*y'+(1+y')/(x+y)-2e^y-2xe^y*y'=0
[1/(x+y)-2sin2y-2xe^y]*y'=2e^y-1/(x+y)
y'=[2e^y-1/(x+y)]/[1/(x+y)-2sin2y-2xe^y]

Answer 3

cos(2y)+ln(x+y)-2x.e^y=0
两边求导
-cos(2y) .(2y') +(1+y')/(x+y)- 2(1+xy').e^y=0
[2x.e^y + 1/(x+y) + 2cos(2y) ].y' = 1/(x+y) -2e^y
y' =[1/(x+y) -2e^y]/[2x.e^y + 1/(x+y) + 2cos(2y) ]

追问

cos2y导数不应该是-sin2y吗

追答

cos(2y)+ln(x+y)-2x.e^y=0
两边求导
-sin(2y) .(2y') +(1+y')/(x+y)- 2(1+xy').e^y=0
[2x.e^y - 1/(x+y) + 2sin(2y) ].y' = 1/(x+y) -2e^y
y' =[1/(x+y) -2e^y]/[2x.e^y - 1/(x+y) + 2sin(2y) ]

Answer 4

将已知方程
cos2y+ln(x+y)-2xeʸ＝0
的两边对x求导得
-2y'sin2y+(1+y')/(x+y)-2eʸ-2xy'eʸ＝0，
两边都乘(x+y)得
-2y'(x+y)sin2y+1+y'-2(x+y)eʸ-2xy'(x+y)eʸ＝0，
移项得
-2y'(x+y)sin2y+y'-2xy'(x+y)eʸ
＝2(x+y)eʸ-1，
提取y'得
y'[1-2(x+y)sin2y-2x(x+y)eʸ]
＝2(x+y)eʸ-1，
提取2(x+y)得
y'[1-2(x+y)(sin2y-xeʸ)]＝2(x+y)eʸ-1，
∴ y'＝[2(x+y)eʸ-1]/[1-2(x+y)(sin2y-xeʸ)].