- #1
footprints
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- 0
[tex]x^2(x+1)(x-2)^7[/tex]
Could someone show me how to start?
Could someone show me how to start?
How to Begin Differentiating Products with x^2(x+1)(x-2)^7?
- Thread starter footprints
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- Differentiation
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Homework Help Overview
The discussion revolves around differentiating the product of three functions: \(x^2\), \((x+1)\), and \((x-2)^7\). Participants are exploring the application of differentiation rules in this context.
Discussion Character
- Exploratory, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- The original poster seeks guidance on how to begin differentiating the product. Some participants suggest combining parts of the product before applying the product rule. Others raise questions about the complexity of differentiation when dealing with larger polynomials and introduce the Leibniz rule for multiple factors.
Discussion Status
Participants are actively engaging with the problem, offering different perspectives on how to approach the differentiation. There is a mix of suggestions regarding the use of the product rule and considerations for more complex cases, indicating a productive exploration of the topic.
Contextual Notes
There is an implicit assumption that the differentiation rules discussed apply to polynomials of varying complexity, with some participants questioning the practicality of direct multiplication in more complicated scenarios.
- #2
Benny
- 577
- 0
Well you currently have a product of three functions. I suggest starting by 'combining' the first two parts of the product as follows.
[tex] \frac{d}{{dx}}\left[ {x^2 \left( {x + 1} \right)\left( {x - 2} \right)^7 } \right] = \frac{d}{{dx}}\left[ {\left( {x^3 + x^2 } \right)\left( {x - 2} \right)^7 } \right][/tex]
Now just use the product rule to differentiate it.
[tex] \frac{d}{{dx}}\left[ {x^2 \left( {x + 1} \right)\left( {x - 2} \right)^7 } \right] = \frac{d}{{dx}}\left[ {\left( {x^3 + x^2 } \right)\left( {x - 2} \right)^7 } \right][/tex]
Now just use the product rule to differentiate it.
- #3
footprints
- 100
- 0
Ah! I thought so to. But just wasn't sure. Thanks! Hapyy new year! 
- #4
dextercioby
Science Advisor
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In this simple case it works,as u're,mutiplying two simple polynomials.But what if thepolynomials had 50 terms?Would u do 2500 multiplications?
Here's the deal:the Leibniz rule is very general.It can be easily extended to finite arbitrary number of factors:
[tex](ABC...Z)'=A'BC...Z+AB'C...Z+ABC'...Z+...+ABC...Z'[/tex]
In your case,there are only 3 very simple polynomials.If u want to,u may not make the multiplications after the differentiation.
Daniel.
EDIT:'Prime' denotes differentiation.
Here's the deal:the Leibniz rule is very general.It can be easily extended to finite arbitrary number of factors:
[tex](ABC...Z)'=A'BC...Z+AB'C...Z+ABC'...Z+...+ABC...Z'[/tex]
In your case,there are only 3 very simple polynomials.If u want to,u may not make the multiplications after the differentiation.
Daniel.
EDIT:'Prime' denotes differentiation.
Last edited:
- #5
footprints
- 100
- 0
Just checking, [tex]A'[/tex] is what I get after I differentiate A right?
Last edited:
- #6
Nylex
- 552
- 2
footprints said:
Yes it is.
- #7
footprints
- 100
- 0
Thanks Nylex! Happy New Year!