Question

设z=In(1+x²+y²),则全微分dz=_____

Answer 1

∂z/∂x=2x/(1+x²+y²)

∂z/∂y=2y/(1+x²+y²)

所以
dz=2xdx/(1+x²+y²)+2ydy/(1+x²+y²)

Answer 2

z=ln(1+x²+y²)
∂z/∂x=2x[1/(1+x²+y²)]
∂z/∂y=2y[1/(1+x²+y²)]
dz=∂z/∂xdx+∂z/∂ydy
= 2x/(1+x²+y²)dx + 2y/(1+x²+y²)dy
dz = 2(xdx + ydy)/(1+x²+y²)