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 Calculus Calculus Math Forum

 January 31st, 2016, 01:04 PM #1 Member   Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 Differentiation using Chain Rule Hi Guys, I am stuck at a question and wondering if anyone can give me some feedback as to what I'm doing wrong. Question: y = 5(x+3)^2(2x-1)^5 My Solution: = 5(2)(x+3)(1)(2x-1)^5 + 5(x+3)^2[5(2x-1)^4(2)] = 10(x+3)(2x-1)^5 + 5(x+3)^2 10(2x-1)^4 = 10(x+3)(2x-1)^5 + 50(x+3)^2(2x-1)^4 However, my answer is nowhere close to the answer in the book and I just can't figure out what I am doing wrong. The answer in the book is 70(x+3)(2x-1)^4(x+2) Please help. Last edited by skipjack; January 31st, 2016 at 01:26 PM. January 31st, 2016, 01:30 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 You're absolutely correct, you just haven't factorised. There's a common factor of $10(x + 3)(2x - 1)^4$. \begin{align*} &10(x + 3)(2x - 1)^5 + 50(x + 3)^2(2x - 1)^4\\\\ =&10(x + 3)(2x - 1)^4(2x - 1 + 5(x + 3))\\\\ =&10(x + 3)(2x - 1)^4(7x + 14)\\\\ =&70(x + 3)(2x - 1)^4(x + 2) \end{align*} Thanks from atari January 31st, 2016, 01:33 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Your answer = 10(x + 3)(2x - 1)^4((2x - 1) + 5(x + 3)) = the book's answer. Thanks from atari January 31st, 2016, 01:54 PM   #4
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Quote:
 Originally Posted by Azzajazz You're absolutely correct, you just haven't factorised. There's a common factor of $10(x + 3)(2x - 1)^4$. \begin{align*} &10(x + 3)(2x - 1)^5 + 50(x + 3)^2(2x - 1)^4\\\\ =&10(x + 3)(2x - 1)^4(2x - 1 + 5(x + 3))\\\\ =&10(x + 3)(2x - 1)^4(7x + 14)\\\\ =&70(x + 3)(2x - 1)^4(x + 2) \end{align*}
That's great, thank you so much. But would you be kind enough to explain how you factored?

I am sorry, I'm taking calculus after 10 years and my basic math skills are quite rusty.

Last edited by skipjack; January 31st, 2016 at 02:10 PM. January 31st, 2016, 02:10 PM #5 Member   Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 Never mind I get it  January 31st, 2016, 03:10 PM #6 Member   Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 Hi Guys, One last question, I solved the question below and wondering if I did it right. Should I simplify by multiplying each terms or should I just leave it the way it is. f(t) = -3(t+9)³(7t²-t) f'(t) = -3(3)(t+9)²(1)(7t²-t) + [-3(t+9)³(14t-1)] = -9(t+9)²(7t²-t)-3(t+9)³(14t-1) = -3(t+9)²(3(7t²-t)+(t+9)(14t-1)) January 31st, 2016, 04:05 PM #7 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond That's correct. Expanding and collecting like terms with 3(7t² - t) + (t + 9)(14t - 1) gives a quadratic in three terms. Thanks from atari Tags chain, differentiation, rule 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 szz Calculus 2 October 29th, 2014 11:22 PM vman0001 Calculus 1 December 24th, 2013 05:43 AM mmmmxxx Calculus 1 March 1st, 2012 12:26 PM Peter1107 Calculus 1 September 8th, 2011 10:25 AM mia6 Calculus 1 April 8th, 2010 07:22 AM

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