Question

设z=y+f(u),u=x²-y²,其中函数f可微,求dz

Answer 1

∂z/∂x=∂f/∂u*∂u/∂x=∂f/∂u*2x=2xf'(u) , 这里记f'(u)=∂f/∂u
∂z/∂y=1+∂f/∂u*∂u/∂y=1+∂f/∂u*(-2y)=1-2yf'(u)
因此dz=∂z/∂xdx+∂z/∂ydy=2xf'(u)dx+[1-2yf'(u)]dy