z=e^(x2+y2)求全微分。求过程

2025-02-28 15:19:00
推荐回答(3个)
回答1:

∂z/∂x=e^(x²+y²)*(x²+y²)'=2xe^(x²+y²)

∂z/∂y=e^(x²+y²)*(x²+y²)'=2ye^(x²+y²)

所以dz=2xe^(x²+y²)dx+2ye^(x²+y²)dy

回答2:

z=e^(x2+y2)

dz=2xe^(x2+y2)dx+2ye^(x2+y2)dy

回答3:

dz = e^(x²+y²)d(x²+y²) = e^(x²+y²)(2xdx+2ydy) =2e^(x²+y²)(xdx+ydy)