Question

设z=u^2ln v,其中u=xy,v=x^2+y^2求z'x和z'y

Answer 1

u=xy
∂u/∂x = y
∂u/∂y = x
v=x^2+y^2
∂v/∂x = 2x
∂v/∂y = 2y
z=u^2.lnv
∂z/∂x
=u^2. (1/v).∂v/∂x + (lnv). ( 2u). ∂u/∂x
=u^2. (1/v). (2x) + (lnv). ( 2u). (2y)
=u^2. (1/v). (2x) + (lnv). ( 2u). (2y)
=(xy)^2. [ 1/(x^2+y^2) ].(2x) + [ln(x^2+y^2) ] .(2xy) (2y)
=2x^3.y^2 /(x^2+y^2) + 4xy^2. ln(x^2+y^2)

z=u^2.lnv
∂z/∂y
=u^2. (1/v).∂v/∂y + (lnv). ( 2u). ∂u/∂y
=u^2. (1/v).(2y) + (lnv). ( 2u). (2x)
=(xy)^2. [ 1/(x^2+y^2) ].(2y) + [ln(x^2+y^2) ] .(2xy) (2x)
=2x^2.y^3 /(x^2+y^2) + 4x^2.yln(x^2+y^2)