# Struggling with quotient rule question

• Jun 7th 2011, 02:40 AM
Turlock
Struggling with quotient rule question
Doing a practise paper question.

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

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

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

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

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

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

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

$g'(x)=\frac{5+9x-2x^2}{(x^2 -6x+23)^\frac{5}{2 } }$
• Jun 7th 2011, 02:53 AM
abhishekkgp
Quote:

Originally Posted by Turlock
Doing a practise paper question.

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

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

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

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

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

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

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

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