How do you use the quotient rule to differentiate #(2x+1)/(x^2-1)#?

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Answer 1 · 2016-01-20T16:01:19

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

Standard form: # (d)/dx (u/v)=(dy)/(dx)= (v (du)/(dx) - u(dv)/(dx))/v^2#

Given:#color(white)(..) y=(2x+1)/(x^2-1)#

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

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

#(dy)/(dx)= (-2x^2-2x-2)/(x^2-1)^2#

But #-2x^2-2x-2= -2(x^2+x+1)#

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