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

1 Answer
Topscooter
Dec 27, 2015

f'(x) = ((2x-6)sqrt(x-3) - (x^2 - 6x + 9)(1/(2sqrt(x-3))))/(x-3)

Explanation:

Let f(x) = (x^2 - 6x + 9)/sqrt(x-3).

The quotient rule tells us that the derivative of (u(x))/(v(x)) is (u'(x)v(x) - u(x)v'(x))/(v(x)^2). Here, let u(x) = x^2 - 6x + 9 and v(x) = sqrt(x-3). So u'(x) = 2x - 6 and v'(x) = 1/(2sqrt(x-3)).

We now apply the quotient rule.

f'(x) = ((2x-6)sqrt(x-3) - (x^2 - 6x + 9)(1/(2sqrt(x-3))))/(x-3)

Related questions
See all questions in Quotient Rule
Impact of this question
1923 views around the world
You can reuse this answer
Creative Commons License