由x+y+z=根号下xyz,确定z是x,y的函数,求dz
问题描述:
由x+y+z=根号下xyz,确定z是x,y的函数,求dz
答
dx+dy+dz=1/[2根号(xyz)]*(yzdx+xzdy+xydz),整理解出dz就可以了
答
d(x+y+z)=d√(x+y+z)dx+dy+dz=1/2√(xyz) d(xyz)dx+dy+dz=1/2√(xyz) (yzdx+xzdy+xydz)(1-xy/(2√xyz))dz=[yz/(2√xyz)-1]dx+[xz/(2√xyz)-1]dydz=[yz/(2√xyz)-1]/(1-xy/(2√xyz))dx+[xz/(2√xyz)-1]/(1-xy/(2√x...