Question

Z”xy＝Z”yx，求Z=ln(y³+ln²x)的二阶偏导数。写一下详细过程

Answer 1

z=ln[y^3+(lnx)^2]
z'x
={ 1/[y^3+(lnx)^2] } . ∂/∂x [y^3+(lnx)^2]
={ 1/[y^3+(lnx)^2] } . [2(lnx)].∂/∂x (lnx)
={ 1/[y^3+(lnx)^2] } . [2(lnx)].(1/x)
=2(lnx)/{ x[y^3+(lnx)^2]}
z''xy
=∂/∂y (z'x)
=∂/∂y [ 2(lnx)/{ x[y^3+(lnx)^2]} ]
=[2(lnx)/x] ∂/∂y { 1/[y^3+(lnx)^2] }
=[2(lnx)/x] .{ -1/[y^3+(lnx)^2]^2 }. ∂/∂y [y^3+(lnx)^2]
=[2(lnx)/x] .{ -1/[y^3+(lnx)^2]^2 }.(3y^2)
=-6(lnx).y^2/{ x. [y^3+(lnx)^2]^2}
//
z'y
= {1/[y^3+(lnx)^2]} . ∂/∂y [y^3+(lnx)^2]
= {1/[y^3+(lnx)^2]} . [3y^2]
=3y^2/[y^3+(lnx)^2]
z''yx
=∂/∂x (z'y)
=∂/∂x { 3y^2/[y^3+(lnx)^2] }
=3y^2. ∂/∂x { 1/[y^3+(lnx)^2] }
=3y^2. { -1/[y^3+(lnx)^2]^2 }.∂/∂x [y^3+(lnx)^2]
=3y^2. { -1/[y^3+(lnx)^2]^2 }. [2(lnx)]. ∂/∂x(lnx)
=3y^2. { -1/[y^3+(lnx)^2]^2 }. [2(lnx)]. (1/x)
=-6(lnx).y^2/{ x.[y^3+(lnx)^2]^2 }
=z''xy