设Z=X2+y2,其中函数y=f(x)由方程X2+y2-xy=1所确定，求dz/dx

Answer 1

dz=2xdx+2ydy, 2xdx+2ydy-ydx-xdy=0 dy=[(y-2x)/(2y-x)]dx
dz=[2x+(2y^2-4xy)/(2y-x)]dx=[(2y^2-2x^2)/(2y-x)]dx