Question

请问这个高数题怎么做？

Answer 1

设u=(x-1)^2=x^2-2x+1, v=x(x^2+1)=x^3+x, w=v/u. 则
u'=2x-2, v'=3x^2+1, w'=(uv'-vu')/u^2, 因此
y'=w'/(3倍三次根号w^2)=(uv'-vu')/(3倍u^2三次根号w^2)
=[(x-1)^2(3x^2+1)-(x^3+x)(2x-2)]/(3倍(x-1)^4三次根号w^2)
=[(x-1)(3x^2+1)-2(x^3+x)]/[3倍(x-1)^3三次根号[x^2(x^2+1)^2/(x-1)^4]]
=[(x-1)(3x^2+1)-2(x^3+x)]/[3倍(x-1)^2三次根号[x^2(x^2+1)^2/(x-1)]]
=(x^3-3x^2-x-1)三次根号(x-1)/[3倍(x-1)^2三次根号[x^2(x^2+1)^2]]
所以dy=(x^3-3x^2-x-1)三次根号(x-1)/[3倍(x-1)^2三次根号[x^2(x^2+1)^2]] dx.

Answer 2

如图

Answer 3

Answer 4

y^3 = x(x^2+1)/(x-1)^2
dy^3 =3y^2 dy = [(3x^2 +1)(x-1)^2 -2(x-1)(x^3+x)]/(x-1)^4dx
=(3x^4 -6x^3 +3x^2 +x^2 -2x+1 -2x^4 -2x +2x^3 +2x)/(x-1)^4dx
=(x^4 -4x^3 +2x^2 -2x +1)/(x-1)^4dx
=(x-1)(x^3 -3x^2 - x -1)/(x-1)^4dx
=(x^3-3x^2 -x-1)/(x-1)^3 dx
dy =(x^3-3x^2 -x-1)/[3(x-1)^3y^2] dx

Answer 5

看不清…………