Question

If f(x)=(2+x)^2(1-x)^3, find f''(x)

I assume we could get the first derivative using the Product Rule:
f'(x)=(2+x)^2(3)(1-x)^2(-1) + (1-x)^3(2)(2+x)(1)
f'(x)=(-3)(2+x)^2(1-x)^2 + (2)(2+x)(1-x)^3
f'(x)=(2+x)(1-x)^2 [-3(2+x) + 2(1-x)]
f'(x)=(2+x)(1-x)^2 [-6-3x + 2-2x]
f'(x)=(2+x)(1-x)^2(-5x-4)

Do correct me if I got it wrong, but I think that's how it is. I need help finding the second derivative. Thanks.

Answer 1

1st derivative is fine, just going through the 2nd.

Answer 2

For finding the second derivative, you need to apply Product Rule twice..like this:
Hint: \[[A(x)*B(x)*C(x)]\prime =A \prime(x)[B(x)*C(x)]+[B(x)*C(x)]\prime A(x)\]
now apply Product Rule again on \[B(x)*C(x)\]..

Answer 3

makes sense?

Answer 4

http://www.wolframalpha.com/input/?i=d%2Fdx%5B%282%2Bx%29%5E2%281-x%29%5E3%5D%3D%282%2Bx%29%281-x%29%5E2%28-5x-4%29

You're right for the first derivative, good job. Use wolfram alpha as a tool and not to cheat and you will be able to come out of calculus with a much stronger understanding.

http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2%5B%282%2Bx%29%5E2%281-x%29%5E3%5D

Answer 5

@Kainui won't it be better if he/she gets to learn how to find that himself using his lil brainy machine? :P

Answer 6

I'm saying exactly that stgreen. But he's looking for confirmation if he's right or not, I believe he understands the method. WA gives you the ability to have instant feedback that homework doesn't supply.

Answer 7

*she lol

Answer 8

lol ohky i see

Answer 9

Yup. I do get that. I could juggle it up, right (like any of them could be A(x), B(x) or C(x))?

I want A(x) to be \[(1-x)^{2}\]
B(x) to be (2+x)
C(x) to be (-5x-4)

So that means I have to do it like:
[A(x)∗B(x)∗C(x)]′=A′(x)[B(x)∗C(x)]+[B(x)∗C(x)]′A(x)
\[f\prime = (2+x)(1-x)^{2}(-5x-4)\] is equal to
\[f\prime\prime = ((1-x)^{2})\prime [(2+x)(-5x-4)] + [(2+x)(-5x-4)]\prime (1-x)^{2}\]
\[f \prime \prime = (2)(1-x)(-1)[(2+x)(-5x-4)] + [(2+x)(-5x-4)]\prime(1-x)^{2}\]
\[-2(1-x)(-5x ^{2}-14x-8) + (-10x-14)(1-x)^{2}\]

Should I continue or am I getting this wrong?

Answer 10

doin' right kido carry on

Answer 11

- sign at the beginning of the last line must be a typo i guess

Answer 12

Hrmm... checked on that. Nope. I didn't write it by mistake.
\[((1-x)^{2})\prime = (2)(1-x)(-1) = -2(1-x)\]
Or did I get it wrong? I'm bad at math. Sorry. :((

Answer 13

no that's alright..you just forgot to put a = sign before it..

Answer 14

I'll be honest. I don't know the next step. Should I be like, looking for a common factor?

Answer 15

wait a min..you wrote last term incorrect in the last line...where is [(2+x)(-5x-4)]' term?

Answer 16

Now all you gotta do is: Start from the second last line and apply product rule on [(2+x)(-5x-4)]' leaving everything else undisturbed

Answer 17

Oops. Multiplied 'em both. So... I should get them back?

Answer 18

well you didn't multiply them well..is that (-10x-14) you got by multiplying (2+x)(-5x-4)?

Answer 19

Multiplied em then derived. Imma start over.

Answer 20

ok lemme fix it (2+x)(-5x-4)=-10x-8-5x^2-4x=-5x^2-14x-8
so you gotta take derivative of this (-5x^2-14x-8)
(-5x^2-14x-8)'=-10x-14
oh yeah that's right i though you just multiplied and didn't differentiate...
anyway last line is GOOD..continue simplifying

Answer 21

got it?you can take (1-x) as common factor from the last line

Answer 22

Cool. I thought I was gonna need to redo the thing.
So starting from:
\[f \prime \prime = -2(1-x)(-5x ^{2}-14x-8) + (-10x-14)(1-x)^{2}\]
I'll factor out (1-x):
\[f \prime \prime = (1-x)[-2(-5x ^{2}-14x-8) + (-10x-14)(1-x)]\]\[f \prime \prime = (1-x)[(10x ^{2}+28x+16) + (-10x-14)(1-x)]\]
Am I still doing this right? :((

Answer 23

trust yourself kid...you're right

Answer 24

I'll continue off:
\[f \prime \prime = (1-x)[(10x ^{2}+28x+16)+(10x ^{2}+4x-14)]\]\[f \prime \prime = (1-x)(20x ^{2}+32x+2)\]
Then, if I multiply 'em both, I'll end up with:
\[-20x ^{3}-12x ^{2}+30x+2\]

Answer 25

Is it done now? :))

Answer 26

sure indeed...you did it :)

Answer 27

Thank you very much, Sir.

Answer 28

Sir?sounds cool..but call me Green :)

Answer 29

Ok, Green. Thanks for helping me out on my homework! :))