How do you differentiate $y=(x+1)/(x^3+x-2)$ ?

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Answer 1 · 2018-05-29T10:33:15

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$y = (x+1)/(x^3+x-2)$

$(dy)/(dx) = ((x^3+x-2)(1)-(x+1)(3x^2+1))/(x^3+x-2)^2$

$(dy)/(dx) = (x^3+x-2-3x^3-x-3x^2-1)/(x^3+x-2)^2$

$(dy)/(dx) = (-2x^3-3x^2-3)/(x^3+x-2)^2$

The quotient rule is given by:

$y=u/v$

$(dy)/(dx) = (vu'-uv')/v^2$

In this case, your u is $x+1$ and your v is $x^3+x-2$

How do you differentiate y=(x+1)/(x^3+x-2)y=(x+1)/(x^3+x-2)?