Question

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

Answer 1

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

Explanation:

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`