第1个回答 2020-07-18
设u=x^3-1, v=x+2, w=uv, r=x+3, s=x^2+6, t=rs.
则u'=3x^2, v'=1, w'=u'v+v'u=3x^2(x+2)+(x^3-1).
r'=1, s'=2x, t'=r's+s'r=(x^2+6)+2x(x+3).
y'=(t'w-w't)/w^2={[(x^2+6)+2x(x+3)](x^3-1)(x+2)-[3x^2(x+2)+(x^3-1)](x+3)(x^2+6)}/[(x^3-1)(x+2)]^2，化简变形检验就有答案了。
则u'=3x^2, v'=1, w'=u'v+v'u=3x^2(x+2)+(x^3-1).
r'=1, s'=2x, t'=r's+s'r=(x^2+6)+2x(x+3).
y'=(t'w-w't)/w^2={[(x^2+6)+2x(x+3)](x^3-1)(x+2)-[3x^2(x+2)+(x^3-1)](x+3)(x^2+6)}/[(x^3-1)(x+2)]^2，化简变形检验就有答案了。
第2个回答 2020-04-29
用对数求导法。
第3个回答 2020-04-29
第4个回答 2020-04-29
y=(x^3-1)(x+2)/(x+3)(x^2+6)
lny=ln(x^3-1)+ln(x+2)-ln(x+3)-ln(x^2+6)
y'/y=(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)
y'=y*[(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)]
=[(x^3-1)(x+2)/(x+3)(x^2+6)]*[(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)]
答案选A本回答被网友采纳
lny=ln(x^3-1)+ln(x+2)-ln(x+3)-ln(x^2+6)
y'/y=(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)
y'=y*[(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)]
=[(x^3-1)(x+2)/(x+3)(x^2+6)]*[(3x^2)/(x^3-1)+1/(x+2)-1/(x+3)-2x/(x^2+6)]
答案选A本回答被网友采纳
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