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

1 Answer
Apr 18, 2017

f'(x)=(2x^2-8x+4)/(x^2-4x+4)

Explanation:

The Quotient Rule tells us that
if f(x)=(g(x))/(h(x))
then
color(white)("XXX")f'(x)=(g'(x) * h(x) - g(x) * h'(x))/(h^2(x))

Since we are told f(x)=(2x^2-3x+2)/(x-2)
it would seem reasonable to let
color(white)("XXX")g(x)=2x^2-3x+2
and
color(white)("XXX")h(x)=x-2

Using the exponent rule for derivatives, this implies
color(white)("XXX")g'(x)=4x-3
and
color(white)("XXX")h'(x)=1

Furthermore, by simple multiplication:
color(white)("XXX")h^2(x)=x^2-4x+4

Evaluating one term at a time for the numerator:
{: (g'(x) * h(x)=,color(white)("xxx"),4x^2-11x+6), (ul(g(x) * h'(x)=),,ul(2x^2-3x+2)), (g'(x) * h(x) - g(x) * h'(x)=,,2x^2-8x+4) :}

f'(x)=(2x^2-8x+4)/(x^2-4x+4)