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 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: 20,379 Thanks: 2011 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: 7,923 Thanks: 1123 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

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