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OpenStudy (anonymous):
(d/dx)[(x+8)/(x^2+x+2)]=
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OpenStudy (anonymous):
quotient rule again [(x+8)'(x^2+x+2)-(x+8)(x^2+x+2)']/(x^2 + x + 2)^2 [1(x^2+x+2) -(x+8)(2x+1)]/(x^2+x+2)^2 [x^2 + x + 2 - 2x^2 - 17x - 8]/(x^2 + x + 2)^2 (-x^2 - 16x - 6) / (x^2 + x +2 )^2
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