设函数z=ln(1+x2+y2)，求dz|1，2

设函数z=ln(1+x²+y²)，求dz|_1，2
【正确答案】：dz=(∂z/∂x)dx+(∂z/∂y)dy =[2x(1+x²+y²)]dx+[2y/(1+x²+y²)]dy， 故dz|_(1,2)=[(2×1)/(1+1²+2²)]dx+[(2×2)/(1+1²+2²)]dy =(1/3)dx+(2/3)dy.