Question

If f(x)=(x-1)(x^2+2)^3, then f'(x)=?
I did this so far:
u=x-1, u'=1, v=(x^2+2)^3, v'=6x(x^2+2)^2
Now when I use the product rule, I keep getting the wrong answer.

Answer 1

it should work, show me how you worked out the product rule

Answer 2

(x-1)(6x(x^2+1)^2)+(x^2+2)^3
Now I don't know which one to expand or multiply
If i open parentheses, I get (x-1)(x^4+2x^2+1)+(x^6+6x^4+12x^2+8)
=x^5+2x^3+x-x^4-2x^2-1+x^6+6x^4-12x^2+8
=x^6+x^5+5x^4+2x^3-24x^2+x+7

the choices are
A) 6x(x^2+2)^2
B) 6x(x-1)(x^2+2)^2
C) (x^2+2)^2 (x^2+3x-1)
D) (x^2+2)^2 (7x^2-6x+2)
E) -3(x-1)(x^2+2)^2

Answer 3

it is d because
\[\large (x-1)*6x*(x^2+2)^2+(x^2+2)^3=(x^2+2)^2 (7x^2-6x+2)\]

Answer 4

but it's pretty long to write it all down...

Answer 5

so which parentheses did you expand? how did you get 7x^2−6x+2?

Answer 6

first factor out (x^2+2)^2 from both sides

Answer 7

so you'll be left with \( \large (x^2+2)^2((x-1)6x+(x^2+2)^1) \)

Answer 8

then distribute the 6x over x-1

Answer 9

\( \large (x^2+2)^2(6x^2-6x+x^2+2) \)

Answer 10

Okay I got it now. Thank you!

Answer 11

no problem