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

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Answer 1 · 2018-05-29T10:33:15

Below

#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#