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July 22nd, 2014, 08:42 PM   #1
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finding derivative

I need help finding the derivative of f(x)=(x-2)/((x^2-x-1)^2)

book says its -3(x^2-3x+1)/((x^2-x+1)^3)

i understand the quotient rule and finding those derivatives, I don't think I understand the algebra and factoring.

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July 23rd, 2014, 07:26 AM   #2
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I disagree with the "book" answer ... check the original f(x)

$\displaystyle f'(x) = \frac{(x^2-x-1)^2 - (x-2) \cdot 2(x^2-x-1)(2x-1)}{(x^2-x-1)^4}$

$\displaystyle f'(x) = \frac{(x^2-x-1)[(x^2-x-1) - (x-2) \cdot 2(2x-1)]}{(x^2-x-1)^4}
$

$\displaystyle f'(x) = \frac{(x^2-x-1)[(x^2-x-1) - (4x^2-10x+4)]}{(x^2-x-1)^4}
$

$\displaystyle f'(x) = \frac{(x^2-x-1)[-3x^2+9x-5]}{(x^2-x-1)^4}
$

$\displaystyle f'(x) = \frac{-3x^2+9x-5}{(x^2-x-1)^3}
$
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July 23rd, 2014, 08:03 AM   #3
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If the denominator of $\displaystyle f(x)$ is $\displaystyle (x^2-x+1)^2$ instead of $\displaystyle (x^2-x-1)^2$ then the answer in the textbook is correct.
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