How do you find the derivative of #x/(x^2-4)#?

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Answer 1 · 2015-08-11T18:18:58

The answer is in the explanation section.

#y = x/(x^2 - 4)#

#dy/dx# = #(((x^2 - 4) (1)) -((x) (2x)))/(x^2 - 4)^2# [by Quotient Rule]

#dy/dx# = #((x^2- 4) - (2x^2))/(x^2 - 4)^2#

#dy/dx# = #(x^2- 4 - 2x^2)/(x^2 - 4)^2#

#dy/dx# = #(-x^2- 4 )/(x^2 - 4)^2#

Answer 2 · 2015-08-11T20:00:03

Use the product, power and chain rules to find:

#d/(dx) (x/(x^2-4)) = -(x^2+4)/(x^2-4)^2#

#d/(dx) (x/(x^2-4)) = d/(dx) x(x^2-4)^-1#

#=(d/(dx) x)(x^2-4)^-1 + x(d/(dx) (x^2-4)^-1)# [Product Rule]

#=(x^2-4)^-1 - 2x^2(x^2-4)^-2# [Power Rule and Chain Rule]

#=((x^2-4)-2x^2)/(x^2-4)^2#

#=-(x^2+4)/(x^2-4)^2#