Question

求下列方程所确定的隐函数y=y(x)的导数y’或微分dy.

1,exy+ylnx=cos2x,求y’
2，x2+y2+xy=0,求y’
3，xy-ex+ey=1,求dy.
请各位数学高手帮忙求一下上面式子的导数和微分，我对数学真的不懂，越详细越好。1题e后面的xy是次方。谢谢了，我急着用，请大家帮帮忙好吗？

Answer 1

楼上的求错了！

1,令F(x,y) = e^(xy)+ylny-cos2x则可由隐函数存在定理求dy/dx = -F'x/F'y
F'x是F对x的偏导数(把y看成定量，然后对x求导),F'y类似
F'x = ye^(xy)+2sin2x, F'y = xe^(xy)+lny + 1
于是dy/dx = -[ye^(xy)+2sin2x]/[xe^(xy)+lny + 1]
2,F（x,y）=x^2+y^2+xy
F'x = 2x+y , F'y = 2y+x => dy/dx = -(2x+y)/(2y+x)
3,F(x,y) = xy-e^x+e^y-1
=> dy = -(F'x/F'y)dx
=[-(y-e^x)/(x+e^y)]dx

Answer 2

w=exy
exy=eu
dw/dx=(dw/du)*(du/dx)=cxy*(xdy+ydx)
d(ylnx)=y*(1/x)dx+lnx*dy
d(cos)=-sin(x)*dx
cxy*(xdy+ydx)+y*(1/x)dx+lnx*dy=-sin(x)dx
cxy*(xy'+y)+y*(1/x)+lnx*y'=-sin(x)
cxy*xy'+lnx*y'=-(sin(x)+cxy*y+y*y*(1/x))
y'==-(sin(x)+cxy*y+y*y*(1/x))/(cxy*x+lnx)

2. 2x+2y2*y'+x*y'+y=0
y'=-2x/(2y2+x)
3. xdy+ydx-ex*dx+ey*dy=0
dy=(ex-y)/(x+ey)