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(2x1)^5 My Solution: = 5(2)(x+3)(1)(2x1)^5 + 5(x+3)^2[5(2x1)^4(2)] = 10(x+3)(2x1)^5 + 5(x+3)^2 10(2x1)^4 = 10(x+3)(2x1)^5 + 50(x+3)^2(2x1)^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)(2x1)^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*} 
January 31st, 2016, 01:33 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,641 Thanks: 2083 
Your answer = 10(x + 3)(2x  1)^4((2x  1) + 5(x + 3)) = the book's answer.

January 31st, 2016, 01:54 PM  #4  
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0  Quote:
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(7t²t) f'(t) = 3(3)(t+9)²(1)(7t²t) + [3(t+9)³(14t1)] = 9(t+9)²(7t²t)3(t+9)³(14t1) = 3(t+9)²(3(7t²t)+(t+9)(14t1)) 
January 31st, 2016, 04:05 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 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.


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