题目

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.
我对数学真的不懂,越详细越好.1题e后面的xy是次方.

提问时间：2020-08-10

答案

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