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 Calculus Calculus Math Forum

 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. Tags derivative, finding Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MrCones Calculus 1 November 18th, 2012 11:47 AM dunn Calculus 3 October 27th, 2011 04:10 AM wulfgarpro Algebra 10 May 13th, 2010 03:45 AM mt055 Calculus 2 November 6th, 2009 07:17 PM Fad3away Calculus 1 January 15th, 2009 03:46 PM

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