My Math Forum Product rule into chain rule

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 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: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Product rule into chain rule You may clean up your result a bit to write: $f'(t)=7t^6$$t^3+4$$^6+18t^9$$t^3+4$$^5$ Note that both terms have $t^6$$t^3+4$$^5$ as factors, so we may write: $f'(t)=t^6$$t^3+4$$^5$$7\(t^3+4$$+18t^3\)=t^6$$t^3+4$$^5$$25t^3+28$$$
 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: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Product rule into chain rule The 3 and the 6 have a product of 18 and the $t^7$ and the $t^2$ have a product of $t^9$.
 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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