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 July 22nd, 2014, 08:42 PM #1 Newbie   Joined: May 2014 From: Texas Posts: 8 Thanks: 0 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. Thanks
 July 23rd, 2014, 07:26 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,017 Thanks: 1603 I disagree with the "book" answer ... check the original f(x) $\displaystyle f'(x) = \frac{(x^2-x-1)^2 - (x-2) \cdot 2(x^2-x-1)(2x-1)}{(x^2-x-1)^4}$ $\displaystyle f'(x) = \frac{(x^2-x-1)[(x^2-x-1) - (x-2) \cdot 2(2x-1)]}{(x^2-x-1)^4}$ $\displaystyle f'(x) = \frac{(x^2-x-1)[(x^2-x-1) - (4x^2-10x+4)]}{(x^2-x-1)^4}$ $\displaystyle f'(x) = \frac{(x^2-x-1)[-3x^2+9x-5]}{(x^2-x-1)^4}$ $\displaystyle f'(x) = \frac{-3x^2+9x-5}{(x^2-x-1)^3}$
 July 23rd, 2014, 08:03 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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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