How to Apply the Product Rule to Find f ' (X) for f(x) = e^{4x}(1-2x)^4?

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Discussion Overview

The discussion centers on applying the product rule to find the derivative of the function f(x) = e^{4x}(1-2x)^4. Participants are seeking clarification on the simplification process after applying the product rule, as well as the correct application of the rule itself.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in simplifying the derivative after applying the product rule.
  • Another participant confirms the need to find the derivative using the product rule and shares a proposed answer, which is met with requests for a fully worked solution.
  • Some participants discuss the need for algebraic manipulation after applying the product rule to reach the textbook answer.
  • There is mention of Leibniz's rule, with one participant asking for clarification on its application.
  • Participants share different forms of the product rule, with some expressing confusion over the notation used.
  • One participant provides a factorization of the derivative but is met with a correction regarding the simplification process.
  • Another participant acknowledges the clarity of a different representation of the product rule compared to what they were taught.

Areas of Agreement / Disagreement

Participants generally agree on the need to apply the product rule but express differing views on the simplification process and the representation of the rule. There is no consensus on the final form of the derivative or the best approach to simplify it.

Contextual Notes

Some participants note the importance of factoring common terms after applying the product rule, but the specific steps and assumptions involved in the simplification remain unresolved.

Who May Find This Useful

Students learning about differentiation, particularly those studying the product rule and its applications in calculus.

Samael
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A problem which I'm sure is rather simple, however I cannot seem to simplify the equation properly to produce the fully simplified answer as written in my textbook.

The problem being.

[tex]f(x)= e^{4x}{(1-2x)}^4[/tex]

Find f ' (X)


All help is greately appreciated!
 
Last edited:
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Are you supposed to take the derivative?
From the title, you already know you can use the product rule. Show us what answer you got.
 
Yes, find the derivative using the PR. The problem I'm having is that I cannot seem to simplify the problem after applying the rule.

The answer to the above problem was:

[tex]-4e^{4x}{(1-2x)}^3(2x+1)[/tex]

However a fully worked solution would explain what went on.
 
You can't simplify more than identifying the 2 terms in the product and apply Leibniz' rule...

BTW,you'll need some algebra afterwards to get to their answer.

Welcome to Physicsforums and,on behalf of the crew,I thank you for the trust and the
support for this forum.:smile:


Daniel.
 
That algebra means factoring the common pieces in the sum you get.

HINT:the exponential and the polynomial with the degree "3".

Daniel.
 
dextercioby said:
You can't simplify more than identifying the 2 terms in the product and apply Leibniz' rule...



Daniel.

I don't think I've ever heard, or been taught that rule before. Would you be able to elaborate on it?
 
It's the product rule,invented by Gottfried Wilhelm Leibniz around 1780.

[tex](ab)'=a'b+ab'[/tex]

Daniel.
 
Ok. However occording to the rule the problem should become:

[tex]e^{4x} .-8{(1-2x)}^3 + {(1-2x)}^4. 4e^{4x}[/tex]

Where to go from here, I am not so sure.
 
Last edited:
  • #10
Sorry, I placed a totally different problem and had to go back and fix it.
 
  • #11
Here's what I'm getting

[tex]f'(x)=\left(e^{4x}\right)'(1-2x)^{4}+e^{4x}\left[(1-2x)^{4}\right]'<br /> =4e^{4x}(1-2x)^{4}-8e^{4x}(1-2x)^{3}[/tex]

Now do what i said,factor the common parts.

Daniel.
 
  • #12
I'm not familar with that form of the product rule. I was thinking of:
[tex]uv'+vu'[/tex]
 
  • #13
It's not smart to switch between the variables (change their order) in the product.We physicists never do it.

Daniel.
 
  • #14
That is the Rule we have been taught, although your representation of it now makes a whole lot more sense than the method we are told to use. Thanks for that.

Also factorising your equation gave:
[tex](4e^{4x} - 8e^{4X} (1-2x)^3<br /> <br /> =<br /> <br /> -4e^{4x}(1-2x)^3[/tex]
 
Last edited:
  • #15
Nope.

[tex]4e^{4x}(1-2x)^{3}\left[(1-2x)-2\right]=-4e^{x}(1-2x)^{3}(1+2x)[/tex]

Daniel.
 
  • #16
Samael said:
That is the Rule we have been taught, although your representation of it now makes a whole lot more sense than the method we are told to use. Thanks for that.

Also factorising your equation gave:
[tex](4e^{4x} - 8e^{4X} (1-2x)^3<br /> <br /> =<br /> <br /> -4e^{4x}(1-2x)^3[/tex]

Your representations of the product rule are the same.
 
  • #17
Thanks for the assistance, its much appreciated. :)
 

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