求下列复合函数的偏导数,z=u^2lnv其中u=y⼀x,v=x^2+y^2

2025-01-08 14:27:50
推荐回答(1个)
回答1:

z=u^2 *lnv
那么偏导数Z'x=2u *lnv *U'x +u^2 *1/v *V'x

Z'y=2u *lnv *U'y +u^2 *1/v *V'y
而u=y/x,所以U'x= -y/x^2,U'y=1/x
v=x^2+y^2,即V'x=2x,V'y=2y
故得到Z'x=2y/x *ln(x^2+y^2) *(-y/x^2) +y^2 /x^2 *1/(x^2+y^2) *2x
= -2y^2 /x^3 *ln(x^2+y^2) +2y^2 /(x^3+xy^2)
Z'y=2y/x *ln(x^2+y^2) *1/x +y^2 /x^2 *1/(x^2+y^2) *2y
=2y/x^2 *ln(x^2+y^2) +2y^3 /(x^4+x^2 y^2)