Question

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

Answer 1

Quotient Rule:
`d/dx {f(x)/g(x)} = (f'(x)g(x)-f(x)g'(x))/g(x)^2`

Original Equation:
`(x^2+3x-6)/(2x-4)`

First, find the derivative of the numerator:
`2x+3`

Times that by the denominator:
`(2x+3)(2x-4)`
`4x^2-2x-12`
Find the derivative of the denominator:
`2`

Times it by the numerator:
`2(x^2+3x-6)`
`2x^2+6x-12`

Subtract the two:
`4x^2-2x-12-2x^2+6x-12`
`2x^2+4x-24`

Square the denominator and divide by it:
`(2x^2+4x-24)/(2x-4)^2`

You can't simplify, so
`f'(x)=(2x^2+4x-24)/(2x-4)^2`

Now repeat the above steps, but with `f'(x)`:
`((4x+4)(2x-4)^2-(2x^2+4x-24)(2)(2x-4))/(2x-4)^4`
`((4x+4)(2x-4)-(2x^2+4x-24)(2))/(2x-4)^3`
`(8x^2-8x-16-4x^2-8x+48)/(2x-4)^3`
`(4x^2-16x+32)/(2x-4)^3`

`f''(x)=(4x^2-16x+32)/(2x-4)^3`