Question

设z=z(x,y)是有方程x²-z²+ln(y/z)=0确定的函数,求dz

Answer 1

dz=(бz/бx)dx+(бz/бy)dy
由x²-z²+ln(y/z)=0求出бz/бx、бz/бy
1、两边对x求偏导
2x-2z(бz/бx)+(z/y){[0-y(бz/бx)]/z^2}=0
解得бz/бx=[2xz/(2z^2+1)]
2、两边对y求偏导
-2z(бz/бy)+(z/y){[z-y(бz/бy)]/z^2}=0
解得：бz/бy={z/[y(1+2z^2)]
dz=(бz/бx)dx+(бz/бy)dy
=[2xz/(2z^2+1)]dx+{z/[y(1+2z^2)]dy
=[z/(2z^2+1)](2xdx+dy/y)