Question

高数：二阶微分

【高数】
已知：μ=(x²+y²+z²)½，求d²μ≥0，哪位大大帮忙算下啊，算了半天算不出正确答案，正确答案是d²μ=[(xdy-ydx)²+(ydz-zdy)²+(zdx-xdz)²]/μ³≥0

Answer 1

u^2=x^2+y^2+z^2
udu=xdx+ydy+zdz
du=(x/u)dx+(y/u)dy+(z/u)dz
d^2u=d(x/u)dx+d(y/u)dx+d(z/u)dz
d(x/u)=(udx-xdu)/u^2,d(y/u)=...,d(z/u)=...
d^2u=(1/u^2)[(udx-xdu)dx+(udy-ydu)dy+(udz-zdu)dz]
=(1/u^2)[u(dx)^2+u(dy)^2+u(dz)^2-(xdx+ydy+zdz)du]
=(1/u^2)[u(dx)^2+u(dy)^2+u(dz)^2-u(du)du]
=(1/u^3)[u^2(dx)^2+u^2(dy)^2+u^2(dz)^2-(udu)^2]
=(1/u^3)[(x^2+y^2+z^2)((dx)^2+(dy)^2+(dz)^2)-(xdx+ydy+zdz)^2]
=(1/u^3)[(xdy-ydx)²+(ydz-zdy)²+(zdx-xdz)²]