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January 31st, 2016, 01:04 PM   #1
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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.
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January 31st, 2016, 01:30 PM   #2
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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*}
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January 31st, 2016, 01:33 PM   #3
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Your answer = 10(x + 3)(2x - 1)^4((2x - 1) + 5(x + 3)) = the book's answer.
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January 31st, 2016, 01:54 PM   #4
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Quote:
Originally Posted by Azzajazz View Post
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.
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January 31st, 2016, 02:10 PM   #5
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Never mind I get it
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January 31st, 2016, 03:10 PM   #6
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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))
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January 31st, 2016, 04:05 PM   #7
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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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