You need to use the quotient rule for the first term of function such that:

`f'(x) = (2'*(x - 1) - 2(x - 1)')/(x - 1)^2 + (18x)'`

`f'(x) = (0*(x - 1) - 2*(1 - 0))/(x - 1)^2 + 18*1`

`f'(x) = (0 - 2)/(x - 1)^2 + 18`

...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

You need to use the quotient rule for the first term of function such that:

`f'(x) = (2'*(x - 1) - 2(x - 1)')/(x - 1)^2 + (18x)'`

`f'(x) = (0*(x - 1) - 2*(1 - 0))/(x - 1)^2 + 18*1`

`f'(x) = (0 - 2)/(x - 1)^2 + 18`

`f'(x) = -2/(x - 1)^2 + 18`

Bringing the term to a common denominator yields:

`f'(x) = (-2 + 18(x - 1)^2)/(x - 1)^2`

Expanding the binomial `(x - 1)^2` yields:

`f'(x) = (-2 + 18(x^2 - 2x + 1))/(x - 1)^2`

`f'(x) = (-2 + 18x^2 - 36x + 18)/(x - 1)^2`

`f'(x) = (18x^2 - 36x + 16)/(x - 1)^2`

Factoring out 2 yields:

`f'(x) = 2(9x^2 - 18x + 8)/(x - 1)^2`

**Hence, evaluating the simplified form pf derivative of the given function yields `f'(x) = 2(9x^2 - 18x + 8)/(x - 1)^2` .**