# Simplifying derivative after using product rule

1. Feb 28, 2009

### msc8127

1. The problem statement, all variables and given/known data
Find the derivative of the function: f(x) = x^2((x-2)^4)

2. Relevant equations

f(x) = (u)(y) + (u)(y)

3. The attempt at a solution

Via the product rule I got [x^2(4(x-2)^3)] + [(2x)(x-2)^4]

the book then goes on to give the solution as 2x(x-3)^3 * (3x-2)

I'm not seeing how the simplification was done.

2. Feb 28, 2009

### Staff: Mentor

Look for the greatest common factor of the two terms, namely, 2x(x - 2)^3. What you'll have left from the first term is x - 2, and from the second is 2x, or 3x - 2.

So your expression is equal to 2x(x - 2)^3 *(3x - 2).

3. Feb 28, 2009

### Dick

2, x and (x-2)^3 are factors of both terms from your product rule. Factor them out and combine what's left.

4. Feb 28, 2009

### msc8127

Are you able to factor the 2x out of both sides by using the power rule on the x^2 in the terms [x^2(4(x-2)^3)]?

i'm really not sure I understand where the 2x comes from in the first term, or where the 4 goes during the simplification.

Sorry, I'm sure i'm making this more complicated than it is supposed to be.

5. Feb 28, 2009

### Dick

4x^2=(2x)*(2x), yes?

6. Feb 28, 2009

### Staff: Mentor

Here's your version of f'(x) (with extraneous parens and brackets removed for clarity): x^2 *4(x-2)^3 + 2x(x-2)^4

2, x, and (x - 2)^3 are factors of both expressions. Bringing 2x(x - 2)^3 out of each expression leaves 2x in the first term, and x - 2 in the second. The final result is
2x(x - 2)^3 * (2x + x - 2)

That last factor simplifies to 3x - 2, and gives your the version in your book.

7. Feb 28, 2009

### msc8127

I see where you all are getting the solution now I think.

The next similar example in the text book is: f(x) = x(3x-7)^3

There is no solution for this problem in the back of the text, so if you all could post a solution for me to check what I get against I would really appreciate it.

thank you all for the help!

8. Feb 28, 2009

### Staff: Mentor

9. Feb 28, 2009

### msc8127

no problem!

Here's what I did:

f(x) = x(3x-7)^3

So via product rule I get:

f(x) = (x)(3)(3x-7)^2 + (1)(3x-7)^3

= (3x-7)^2 * (6x-7)

With my luck something there isn't correct, but that's how I worked the problem.

Thank you!

10. Feb 28, 2009

### yyat

Almost. When you differentiate (3x-7)^3 you need to apply the chain rule, giving you an extra factor of 3=(3x-7)'.

11. Feb 28, 2009

### msc8127

ok, using the chain rule I get f(x) = [(x)(3)(3x-7)^2] +(3x-7)^3

which if I take out (3x-7)^2 from both terms, I am left with (3x-7)^2 * [(3x)*(3x-7)]

From there I can take out a 3x from inside the brackets and get:

[(3x)*(3x-7)^2] * (-7) = f(x)??

12. Feb 28, 2009

### yyat

No! As I tried to explain before, the derivative of (3x-7)^3 is not 3(3x-7)^2 but 9*(3x-7)^2. This is just the chain rule.

I hope this is a typo...

13. Feb 28, 2009

### msc8127

argh...

trying again from the chain rule:

(x)[9(3x-7)^2] + (3x-7)^3 = (3x-7)^2 * (12x-7)

If that's not correct, then i can't figure out where i'm screwing up at. I do see where I was screwing up by not multiplying by the derivative of the "inside" when using the chain rule now....sorry I missed that earlier.

14. Feb 28, 2009

### Dick

Don't fret. Now it's correct.

15. Feb 28, 2009

### msc8127

excellent...Thank you all for guiding my fumbles through the simplifications.

I'll rework these examples a few times and make sure they're in my head the right way.

Thanks again for the help

16. Feb 28, 2009

### msc8127

ok, i've worked through the problems we've went over so far and though I had everything figured out, then another curve ball got me.

the problem i'm working on now is: y= x[(1-x2)1/2]

I applied the product rule and applied the chain rule to get to:

y`= x[(1/2)(1-x2)-1/2(-2x)] + (1-x2)1/2)(1)

I know from the online solutions guide that the next step is -x2(1-x2)-1/2 + (1-x2)1/2

BUT, I'm not sure where the -x2 comes from and where the -2x goes.

am I correct that the 1/2 * -2x gives a -1x which when multiplied by x already outside the parenthesis gives me a -x2?

If so, then I'm in good shape, if not, i'm about to quit college again and go back to mowing yards.

Thanks for the help!

17. Feb 28, 2009

### Dick

Yes, x*(1/2)*(-2x)=(-x^2). Why are you getting so nervous about this stuff? Just do it. Though there is also nothing wrong with mowing lawns.

18. Mar 1, 2009

### msc8127

i'm not sure why i'm letting this stuff get the best of me. There are other topics later in the same chapter that should be more complex, but for some reason this simplification stuff has had my number.

on a side note, I quit college in 2003 (1 semester from Music Ed degree) and have been in the landscaping industry since then. I returned to college in the fall of 2007 having switched to Physics as my field of study. While I do well conceptually with physics, the mathematics side of things really seems to intimidate me. I just keep plugging away at it though, so hopefully there will come a point when things get less stressful.

Thanks for the help!