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

[Samael]

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.

$$f(x)= e^{4x}{(1-2x)}^4$$

Find f ' (X)


All help is greately appreciated!

 

[Galileo]

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.

 

[Samael]

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:

$$-4e^{4x}{(1-2x)}^3(2x+1)$$

However a fully worked solution would explain what went on.

 

[dextercioby]

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.

Daniel.

 

[dextercioby]

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

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

Daniel.

 

[Samael]

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?

 

[dextercioby]

It's the product rule,invented by Gottfried Wilhelm Leibniz around 1780.

$$(ab)'=a'b+ab'$$

Daniel.

 

[Samael]

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

$$e^{4x} .-8{(1-2x)}^3 + {(1-2x)}^4. 4e^{4x}$$

Where to go from here, I am not so sure.

 

[dextercioby]

Fix the LaTex code.

Daniel.

 

[Samael]

Sorry, I placed a totally different problem and had to go back and fix it.

 

[dextercioby]

Here's what I'm getting

$$f'(x)=\left(e^{4x}\right)'(1-2x)^{4}+e^{4x}\left[(1-2x)^{4}\right]' =4e^{4x}(1-2x)^{4}-8e^{4x}(1-2x)^{3}$$

Now do what i said,factor the common parts.

Daniel.

 

[Samael]

I'm not familar with that form of the product rule. I was thinking of:
$$uv'+vu'$$

 

[dextercioby]

It's not smart to switch between the variables (change their order) in the product.We physicists never do it.

Daniel.

 

[Samael]

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:
$$(4e^{4x} - 8e^{4X} (1-2x)^3 = -4e^{4x}(1-2x)^3$$

 

[dextercioby]

Nope.

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

Daniel.

 

[whozum]

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:
$$(4e^{4x} - 8e^{4X} (1-2x)^3 = -4e^{4x}(1-2x)^3$$

Your representations of the product rule are the same.

 

[Samael]

Thanks for the assistance, its much appreciated. :)