Question

f'' of : f(x)= (1-3x)^3 (2x-3) Please :)

Answer 1

so what's the problem here?

Answer 2

I can teach you the power rule, but first I want you to show me how you would do this using the derivative definition

Answer 3

I can't arrive to the answerrrrr

Answer 4

tic toc tic toc tic toc

Answer 5

hey, nin! what's up, man? do you need help in math?

Answer 6

I guess you don't need help anymore @aml13 ?

Answer 7

no yesss

Answer 8

I think the reason you're having a little trouble arriving at the correct answer is because you didn't simplify your function first

Answer 9

but I mean it should be still the same utilizing the power rule
\[f'(x)= (nx^{n-1})\]

Answer 10

Looks like a mistake in the cain rule.

Answer 11

chain

Answer 12

so let us say we have
f(x) = 2x^4

it's derivative would be
f'(x) = 4(2x^3)
= 8x^3

Answer 13

yes but that's what I did

Answer 14

Hmmm...

Answer 15

i'm doing f''

Answer 16

Kk. Then... how did you get that? It looks like the mitake is there.

Answer 17

Take each side of the + seperately as you do that part.

Answer 18

yeah what mick said, mistake with product rule

Answer 19

its not u'v+uv' ??

Answer 20

product
(-3+2 x) (d/dx((1-3 x)^3))+(1-3 x)^3 (d/dx(-3+2 x))

Answer 21

f(x)= (1-3x)^3 (2x-3)

get the first derivative..
f'(x) = 2(1-3x)^2(2x-3) + (1-3x)^3(2)
f''(x) = 4(1-3x)(2x-3)(2-6x)^3) + 2(1-3x)^2(2x-3)(3(2-6x)^2)

Answer 22

It is u'v+uv', but that is not what you did.

Answer 23

its messy so i might've made a mistake..

Answer 24

I had to write it down laughing out loud

Answer 25

I think i did it right lol

Answer 26

haha that's good :)

Answer 27

Do u want to know the real answer?

Answer 28

f''= 18(1-3x)(12x-11)

Answer 29

oh oops i used the product rule again

Answer 30

no wonder LOL

Answer 31

>.<

Answer 32

still need help ?

Answer 33

f'(x) = 2(1-3x)^2(2x-3) + (1-3x)^3(2)
f''(x) = 4(1-3x)(2x-3) + 6(1-3x)^2

Answer 34

And Nin, was the other way you were hinting at the multiply it all out method?

Answer 35

why are you using 2 sham?

Answer 36

laughing out loud mic, that was dirty and loads of work

Answer 37

Hehe. Well... it does "simplify" the problem.

Answer 38

the derivative of (2x-3)
is just 2..

Answer 39

LAUGHING MY ARSE OFF get out

Answer 40

d of u is 3(-3x+1)^2

Answer 41

or did I write it wrong?

Answer 42

and mic, I did it with "simplifying" and it is easier

Answer 43

it expanded it a little bit, but it is manageable and just follows one rule

Answer 44

try it

Answer 45

so , for the f'' do I make de (u'v + uv') rule??

Answer 46

uhh?

Answer 47

so complicated

Answer 48

Okay. i'm going to start over...
f(x) = (1-3x)^3(2x-3)
using the product rule, where u = (1-3x)^3 and v = 2x-3
you get
\[f'(x) =(2x-3) \frac{ d }{ dx }[(1-3x)^3] + (1-3x)^3 \frac{ d }{ dx } (-3+2x)\]

Answer 49

\[\frac{ d }{ dx }( u~v) = v\frac{ du }{ dx } + u \frac{ dv }{ dx }\]

Now using the chain rule..
\[\frac{ d }{ dx } (1-3x)^3 = \frac{ du^3 }{ du } \frac{ du }{ dx }\]
where u = 1-3x, and d/du (u^3) = 3u^2
\[f'(x) = (1-3x)^3 (\frac{ d }{ dx } (-3 + 2x)) + (-3+2x) ~~3(1-3x)^2\frac{ d }{ dx } (1-3x)\]

Answer 50

okay thanks :):)

Answer 51

damn... laughing out loud

Answer 52

nin wot

Answer 53

I think she already got the answer

Answer 54

LOL

Answer 55

No I'm just tired to do this fuging number haha :p..

Answer 56

Finally got it ! hahaha

Answer 57

Yes, that way worls well. There are a few you can use to get there, but that is a good choice.