Question

need help. find the first derivative (1-2x)^2(x-3). How am i to go about it, do i need to expand the first bracket.

Answer 1

you can expand the first bracket if you want. I would not expand and use the chain rule on it instead.

Answer 2

i would expand the whole thing
\[(1-2x)^2(x-3)=4x^2-16x^2+13x-3\] and then it is real easy

Answer 3

i used the chain rule as well, could u please help verify the answer, this is what i got -4(1-2x)(x-3)+(1-2x)^2. i will xpand it and c if i get d same answer.

Answer 4

I was going to say expanding is fine too...ill check

Answer 5

if you do it that way you have to use the product rule too

Answer 6

should all work out the same since all the short cut rules are consistent
but for my money if you are going to have to expand anyway, why not to it first?

Answer 7

@satellite73 agreed

Answer 8

Yeah I get the same

Answer 9

no way.... u cannot get the same when u expand the whole thing. i expand n got 13-16x-12x^2-3

Answer 10

well if u got the same as well then how can i go about findind the second derivative.

Answer 11

better to expand the original function first then differentiate twice

Answer 12

easier to deal with

Answer 13

so please verify my answer when i expanded the whole equation 13-16x-12x^2-3

Answer 14

(1-2x)^2(x-3)=\[(4x^2-4x+1))(x-3)=4x^3-12x^2-4x^2+12x+x-3\]\[4x^3-16x^2+13x-3\]

Answer 15

now differentiate

Answer 16

\[f'(x)=12x^2-32x+13\]

Answer 17

\[f''(x)=24x-32\]

Answer 18

-4x^3+8x^2+13x-3

Answer 19

thats my ans. for first derivative.

Answer 20

where do you get -4x^3

Answer 21

if u expand (1-2x)^2 you will get 1-4x-4x^2 ryt?

Answer 22

wrong. 4x^2-4x+1

Answer 23

let me go bck again

Answer 24

ok

Answer 25

(-2x)^2=4x^2

Answer 26

well well well i got 12x^2-32x+13. as the 1st derivative

Answer 27

is dat correct?

Answer 28

hello eseidi are u ther?

Answer 29

yep, that should be right.