第1个回答 2022-02-20
z = xf(y,x/y)
∂z/∂x = f(y,x/y) + xf'2(y,x/y)(1/y) = f(y,x/y) + (x/y)f'2(y,x/y)
∂z/∂y = x[f'1(y,x/y) + f'2(y,x/y)(-x/y^2)] = x[f'1(y,x/y) - (x/y^2)f'2(y,x/y)]
dz = (∂z/∂x)dx + (∂z/∂y)dy
= [f(y,x/y)+(x/y)f'2(y,x/y)]dx + x[f'1(y,x/y)-(x/y^2)f'2(y,x/y)]dy