# Application of the Product Rule

1. Jun 1, 2005

### 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!

Last edited: Jun 1, 2005
2. Jun 1, 2005

### 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.

3. Jun 1, 2005

### 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.

4. Jun 1, 2005

### 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.

5. Jun 1, 2005

### dextercioby

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

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

Daniel.

6. Jun 1, 2005

### Samael

I don't think I've ever heard, or been taught that rule before. Would you be able to elaborate on it?

7. Jun 1, 2005

### dextercioby

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

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

Daniel.

8. Jun 1, 2005

### 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.

Last edited: Jun 1, 2005
9. Jun 1, 2005

### dextercioby

Fix the LaTex code.

Daniel.

10. Jun 1, 2005

### Samael

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

11. Jun 1, 2005

### 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.

12. Jun 1, 2005

### Samael

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

13. Jun 1, 2005

### dextercioby

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

Daniel.

14. Jun 1, 2005

### 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.

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

Last edited: Jun 1, 2005
15. Jun 1, 2005

### dextercioby

Nope.

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

Daniel.

16. Jun 1, 2005

### whozum

Your representations of the product rule are the same.

17. Jun 3, 2005

### Samael

Thanks for the assistance, its much appreciated. :)