z=lntan(x/y)求偏导数书上的答案是∂z/∂x=2/ycsc2x/y;∂z/∂y=-2x/y^2csc2x/y,这个答案是错的,还是可以化简成这个答案

Question

z=lntan(x/y)求偏导数
书上的答案是∂z/∂x=2/ycsc2x/y;∂z/∂y=-2x/y^2csc2x/y,这个答案是错的,还是可以化简成这个答案

Answer 1

答案当然是对的,
但化简很麻烦,需要细心一些
∂z/∂x
=1/tan(x/y) * ∂[tan(x/y)]/∂x
而
∂[tan(x/y)]/∂x
=1/[cos(x/y)]^2 *∂[(x/y)]/∂x
=1/y* 1/[cos(x/y)]^2
所以
∂z/∂x
=1/tan(x/y) * ∂[tan(x/y)]/∂x
=1/ { tan(x/y) *y *[cos(x/y)]^2 }
=1/y * 1/[sin(x/y) *cos(x/y)]
=2/y * 1/[2sin(x/y) *cos(x/y)] 显然1/[2sin(x/y) *cos(x/y)]=1/sin(2x/y)=csc2x/y
=2/y *csc2x/y
而
∂z/∂y
=1/tan(x/y) * ∂[tan(x/y)]/∂y
=1/tan(x/y) * 1/[cos(x/y)]^2 *∂[(x/y)]/∂y
=1/tan(x/y) * 1/[cos(x/y)]^2 * (-x/y^2)
= -2x/y^2 * 1/ {tan(x/y) *[cos(x/y)]^2}
= -2x/y^2 *csc2x/y