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**Turlock** Doing a practise paper question.

My answer to an earlier part for using composite rule on $\displaystyle f(x)=(x^2 -6x+23)^\frac{3}{2 } $

I got $\displaystyle \frac{3}{2 }(x^ -6x+23)^\frac{1}{2 } (2x-6) $

Now I need to use quotient rule to differentiate $\displaystyle g(x)=\frac{x-2}{(x^2 -6x+23)^\frac{3}{2 } } $

The working so far is $\displaystyle g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }(1)] - [(x-2)(\frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }(2x-6)) }{[(x^2 -6x+23)^\frac{3}{2 }]^2 } $

$\displaystyle g'(x)=\frac{[(x^2 -6x+23)^\frac{3}{2 }] - [(2x^2 - 10x+12) \frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 }] }{(x^2 -6x+23)^\frac{6}{ 2} } - $

$\displaystyle g'=\frac{(x^2 -6x+23)^\frac{3}{2 } - [(3x^2 -15x +18)(x^2 -6x+23)^\frac{1}{2 } }{ (x^2 -6x+23)^\frac{6}{2 } } $

take $\displaystyle (x^2-6x+23)^{\frac{1}{2}}$ out from the numerator and then cancel this with the denominator

and here I'm stuck with how to reach the answer i have of

$\displaystyle g'(x)=\frac{5+9x-2x^2}{(x^2 -6x+23)^\frac{5}{2 } } $