题目

函数z=f(x,y)由方程xy+yz+zx=1所确定,求fxy" .

提问时间：2020-11-28

答案

z对x的偏导 xy+yz+zx=1 y＋yfx'+z+xfx'＝0 z对y的偏导 x＋z+yfy'＋xfy' ＝0 z对y的偏导 1+fx'+yfxy"+fy'+xfxy" =0 1+(fx'+fy')+(x+y)fxy"=0 由 得:1+fx'+fy' =1-(y+z)/(x+y)-(x+z)/(x+y) =-2z/(x+y) (x+y)fxy"=2z/(x+y) fxy" =2z/(x+y)^2