# HELP! trouble factoring while using the chain rule + the product rule

Discussion in 'Community Discussion' started by macman2790, Feb 28, 2007.

1. ### macman2790 macrumors 6502a

Joined:
Sep 4, 2006
Location:
Texas
#1
I'm having trouble factoring using the chain rule with the product rule. the answer is in the book but i cant figure out how to factor it out correctly. Will someone please walk me through the factoring part, i'm having trouble with every problem like this and i have very many. the problem is:
differentiate:
y = (1+4x)^5 (3+x-x^2)^8

heres what I have so far:

y'= [5(1+4x)^4(4)](3+x-x^2)^8 + [8(3+x-x^2)^7*(1-2x)(1+4x^5)

now here's the answer:4(1+4x)^4(3+x-x^2)^7(17 + 9x - 21x^2)

Thanks in advance for the help.

2. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#2
y = (1+4x)^5 (3+x-x^2)^8

y' = 20(1+4x)^4 (3+x+x^2)^8 + 8(1-2x)(3+x-x^2)^7 (1+4x)^5

= 20(1+4x)^4 (3+x+x^2)^7 (3+x+x^2) + 8(1-2x)(3+x-x^2)^7 (1+4x)^4 (1+4x)

= 4(1+4x)^4 (3+x+x^2)^7 [5 (3+x+x^2) + 2 (1-2x) (1+4x)]

I think the answer follows from here by simplifying what is in [...]

3. ### macman2790 thread starter macrumors 6502a

Joined:
Sep 4, 2006
Location:
Texas
#3
thank you very much, i should have a bit of an easier time simplifying the rest of the 15 product rule + chain rule problems

4. ### macman2790 thread starter macrumors 6502a

Joined:
Sep 4, 2006
Location:
Texas
#4
= 20(1+4x)^4 (3+x+x^2)^7 (3+x+x^2) + 8(1-2x)(3+x-x^2)^7 (1+4x)^4 (1+4x)

= 4(1+4x)^4 (3+x+x^2)^7 [5 (3+x+x^2) + 2 (1-2x) (1+4x)]

does anyone mind telling me why 2(1+4x)^4 is simplifying to 4(1+4x), its just too late for me lol.

5. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#5
?

Not sure where you are looking?

I split the 20 up into 4*5, the 4 is still out front, the 5 is inside the [..]

6. ### macman2790 thread starter macrumors 6502a

Joined:
Sep 4, 2006
Location:
Texas
#6
i just saw it right before you posted, thanks, like i said it's too late for math lol.

7. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#7
It's never too late for math