How do you differentiate #(3x^2 -5x )/ (2x-1)# using the quotient rule?

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Jun 2, 2018

Answer:

#y'=(6x^2-6x+5)/(2x-1)^2#

Explanation:

Using the Quotient rule we get
#y'=((6x-5)(2x-1)-(3x^2-5x)*2)/(2x-1)^2#
and this is

#y'=(12x^2-10x-6x+5-6x^2+10x)/(2x-1)^2#
this is

#y'=(6x^2-6x+5)/(2x-1)^2#

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