Question

对arcsin(x-y)^(-1/2)求偏导数该怎么求

Answer 1

f=arcsin(x-y)^(-1/2)

af/ax=(-1/2)arcsin(x-y)^(-3/2)[arcsin(x-y)]`

=(-1/2)arcsin(x-y)^(-3/2)[1/√(1-(x-y)²)*(x-y)`

=(-1/2)arcsin(x-y)^(-3/2)/[√(1-(x-y)²]

af/ay=(-1/2)arcsin(x-y)^(-3/2)[arcsin(x-y)]`

=(-1/2)arcsin(x-y)^(-3/2)[1/√(1-(x-y)²)*(x-y)`

=(1/2)arcsin(x-y)^(-3/2)/[√(1-(x-y)²]