求复合函数的偏导数 设Z=u^2 lnv , u=y/x, v=x^2+y^2, 求 az/ax ,az/ay

求下列复合函数的偏导数，z=u^2lnv其中u=y/x,v=x^2+y^2

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)

z/x=z/u u/x+z/v v/x
=2ulnv ×1+u²/v × 2xy
=2(x+y)ln(x²y)+2(x+y)²/x

z/y=z/u u/y+z/v v/y
=2ulnv ×1 +u²/v ×x²
=2（x+y）ln(x²y)+(x+y)²/y

az/ax=az/au+au/ax=2ulnv-y/x^2

az/ay=az/av+av/ay=u^2/v+2y
然后再稍微化简一下就行啦！