Question

1,设z=z(x,y) 是由方程 2z-x²y+cos(x-z)=0 所确定的隐函数,求dz

Answer 1

2z-x²y+cos(x-z)=0

两边对x求导

2z'-2xy-(1-z')sin(x-z)=0

那么z对x的偏导就是(2xy+sin(x-y))/(2+sin(x-y))

两边对y求导

2z'-x^2+z'sin(x-z)=0

那么z对y的偏导就是x^2/(2+sin(x-z))

所以dz=(2xy+sin(x-y))/(2+sin(x-y))dx+x^2/(2+sin(x-z))dy