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