My Math Forum Quotient Rule

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 January 29th, 2012, 12:00 AM #1 Senior Member   Joined: Jan 2012 Posts: 100 Thanks: 0 Quotient Rule Hey guys, Looking for help here again, I have the function: F(x): (x^2-4)/(x+1). We are supposed to use the quotient rule. I cannot seem to get the correct answer at the back of my book.. I get: (2x+4)/(x+1)^2 Textbook: 3/(2x+1)^2 What am I doing wrong?
 January 29th, 2012, 12:17 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Quotient Rule I get a result different from yours and from that of your textbook: $\frac{d}{dx}$$\frac{x^2-4}{x+1}$$=\frac{(x+1)(2x)-$$x^2-4$$(1)}{(x+1)^2}=\frac{2x^2+2x-x^2+4}{(x+1)^2}=\frac{x^2+2x+4}{(x+1)^2}$
 January 29th, 2012, 12:21 AM #3 Senior Member   Joined: Jan 2012 Posts: 100 Thanks: 0 Re: Quotient Rule I worked that one out previously...I was not sure..My textbook has been sub-par for providing us with the right answers..I have 100% more confidence in going with yours.
 January 29th, 2012, 12:23 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Quotient Rule The closest I can get to your textbook's result is: $\frac{3}{(x+1)^2}+1$
 January 29th, 2012, 04:06 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,692 Thanks: 976 Math Focus: Elementary mathematics and beyond Re: Quotient Rule (x^2-4)/(x+1) $\frac{d}{dx}\,\frac{x^2\,-\,4}{x\,+\,1}\,=\,\frac{2x(x\,+\,1)\,-\,1(x^2\,-\,4)}{(x\,+\,1)^2}\,=\,\frac{2x^2\,+\,2x\,-\,x^2\,+\,4}{(x\,+\,1)^2}\,=\,\frac{x^2\,+\,2x\,+\ ,4}{(x\,+\,1)^2}$ $=\,\frac{x^2\,+\,2x\,+\,1\,+\,3}{x^2\,+\,2x\,+\,1} \,=\,1\,+\,\frac{3}{(x\,+\,1)^2}\,$

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