已知y=根号下x(x-1)^3/(x^2+1)^3 求DY/DX 麻烦

设x(x-1)^3=u,(x^2+1)^3=v,a=x-1,b=x^2+1,u=xa^3,v=b^3
u'=(xa^3)' =x'a^3+x(a^3)'=(x-1)^3+3x(x-1)^2=a^3+3xa^2
v'=3(x^2+1)^2=3b^2,v'/v=3/b
(u/v)'=(u'v-uv')/v^2=u'/v - u/v * v'/v
=(a^3+3xa^2)/b^3 - (a^3/b^3) * (3/b)
=(a^3b+3xba^2-3a^3)/b^4
=(a^2/b^4)*(ab+3xb-3a)
=(a^2/b^4)*(ab+3xb-3a)
=(a^2/b^4)*[(x-1)(x^2+1)+3x(x^2+1)-3(x-1)]
=(a^2/b^4)*[(x-1)(x^2-2)+3x(x^2+1)]
=(a^2/b^4)*(x^3-x^2-2x+2+3x^3+3x)
=(a^2/b^4)*(4x^3-x^2+x+2)
=(4x^3-x^2+x+2)(x-1)^2/(x^2+1)^4