October 19th, 2012, 01:42 PM  #1 
Senior Member Joined: Mar 2012 Posts: 111 Thanks: 0  Product rule into chain rule
I have this problem: t^7(t^3+4)^6 Here are the steps I have so far, but I don't know where to simplify it even more. Ok I'm going to do the product rule first f'(x)g(x) + f(x)g'(x) (7t^6)(t^3+4)^6 + (t7)(t^3+4)^6' ^^ so I have to use the chain rule on this last one. f'(g(x) times g'(x). ^^ I get 6(t^3+4)^5(3t^2) Ok now I have my f' g f and g' (7t^6)(t^3+4)^6 + (t7)6(t^3+4)^5(3t^2) I have no idea where to go from here in terms of simplifying... 
October 19th, 2012, 01:49 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 464 Math Focus: Calculus/ODEs  Re: Product rule into chain rule
You may clean up your result a bit to write: Note that both terms have as factors, so we may write: 
October 19th, 2012, 01:56 PM  #3 
Senior Member Joined: Mar 2012 Posts: 111 Thanks: 0  Re: Product rule into chain rule
Mark, how did you get the 18t^9(t^3+4)^5 from (t7)6(t^3+4)^5(3t^2)?! You're a genius man lol. 
October 19th, 2012, 02:04 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 464 Math Focus: Calculus/ODEs  Re: Product rule into chain rule
The 3 and the 6 have a product of 18 and the and the have a product of .

October 19th, 2012, 02:21 PM  #5 
Senior Member Joined: Mar 2012 Posts: 111 Thanks: 0  Re: Product rule into chain rule
gotcha thanks!


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