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macman2790

macrumors 6502a
Original poster
Sep 4, 2006
716
1
Texas
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.
 
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 [...]
 
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 [...]

thank you very much, i should have a bit of an easier time simplifying the rest of the 15 product rule + chain rule problems:D
 
= 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.:D
 
= 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.:D
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I split the 20 up into 4*5, the 4 is still out front, the 5 is inside the [..]
 
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