Question

Differentiate: g(x) = (1+4x)^5 * (3 + x - x^2)^8

Answer 1

I got as far as 8 (4x+1)^5 (-2x+1) (-x^2+x+3)^7 + 20 (4x+1)^4 (-x^2+x+3)^8, but the textbook's answer is 4(1+4x)^4 (3+x-x^2)^7 (17+9x-21x^2) .....

Answer 2

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

Answer 3

wait that 3x is supposed to be 3. Typing long expressions is conusing.

Answer 4

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

Answer 5

yeah but i have to get that to the textbook answer

Answer 6

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

Answer 7

g'(x)=4(1+4x)^4(3+x-x^2)^7[15+5x-5x^2+2(1+4x-2x-8x^2)]

Answer 8

but how

Answer 9

pull out a 4, (1+4x)^4 and (3+x-x^2)^7, those are all common terms

Answer 10

you mean 4, (1+4x) and (3+x-x^2)?

Answer 11

No, to the 7th power.. one is to the 7th and one is to the 8th so just pull out the whole thing

Answer 12

riight wait so divide both sides by 4 (3+x-x^2)^7 (1+4x)^4 ??