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)#

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