How do you differentiate #(4x^2 -6x + 9 )/ (2x-3)# using the quotient rule?

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Answer 1 · 2016-09-20T05:34:53

Derivative of #(4x^2-6x+9)/(2x-3)# is #(8x(x-3))/(2x-3)^2#

Quotient rule states if #f(x)=(g(x))/(h(x))#

then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Hence as #f(x)=(4x^2-6x+9)/(2x-3)#

#(df)/(dx)=((4xx2x-6)xx(2x-3)-2xx(4x^2-6x+9))/(2x-3)^2#

= #((8x-6)(2x-3)-8x^2+12x-18)/(2x-3)^2#

= #(8x xx2x-3xx8x-6xx2x-6xx(-3)-8x^2+12x-18)/(2x-3)^2#

= #(16x^2-24x-12x+18-8x^2+12x-18)/(2x-3)^2#

= #(8x^2-24x)/(2x-3)^2#

= #(8x(x-3))/(2x-3)^2#