Question

设f(x,y,z)=e∧xyz²,其中z=z(x,y)是由方程x+y+z+xyz=0确定,求fx（0,1,-1

Answer 1

∂f/∂x=[yz^2+2xyz∂z/∂x]e^(xyz^2)

x+y+z+xyz=0 1+∂z/∂x+yz+xy∂z/∂x=0 ∂z/∂x(1+xy)=-(1+yz) ∂z/∂x=-(1+yz)/(1+xy)

f'x(0,1,-1) = (1+0)e^(0)=1

f'x(0.1-1)=1