can someone help me solve this problem ?
let f(x)= (x^3+3x+8)^2
F'(x)=
F'(2)=

- anonymous

can someone help me solve this problem ?
let f(x)= (x^3+3x+8)^2
F'(x)=
F'(2)=

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- katieb

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- hartnn

do you know the chain rule ?

- anonymous

not really sure with the chain rule..

- anonymous

need to learn that then to do a question of this type

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## More answers

- anonymous

yes, that's why I'm trying to learn it here.. if that helps.

- hartnn

let me give you a similar example, if
\(f(x) = (ax^2+bx+c)^3 \)
then by chain rule, \(f'(x)=3(ax^2+bx+c)^2 (ax^2+bx+c)' \\ =3(ax^2+bx+c)^2(2ax+b) \)
got this ?

- hartnn

by definition,
\(f(g(x))' = f'(g(x)) \times g'(x)\)

- anonymous

yeah

- hartnn

so try once for your Q, maybe.. ?

- anonymous

you dont subtract the exponent n-1?

- hartnn

i made the exponent n-1
in the Q, ax^2+bx+c had exponent 3, in the answer i became 2, right ?

- hartnn

for x^n, the derivative is
\((x^n)'=n x^{n-1}\)

- anonymous

okay... sort of there. but then the second equation next to it confused me a little.

- hartnn

yeah, from the definition, i needed to multiply by the derivative of inner function also (g'(x))
so ,i multiplied by derivative of ax^2+bx+c which is 2ax+b

- hartnn

in your Q, you'll multiply by (x^3+3x+8)

- hartnn

**by the derivative of (x^3+3x+8)

- anonymous

so the derivative for F'(x) would be.. (3x^2+3x+8)?

- hartnn

no... you have to first derivate the out function, (x^2), then multiply the derivative of each term of x^3+3x+8
first tell whats the derivative of x^2

- anonymous

2x

- hartnn

so, derivative of (x^3+3x+8)^2 will be 2(x^3+3x+8) *(derivative of x^3+3x+8 ) , right ?
can you differentiate, x^3+3x+8 ?

- anonymous

2x^2+6x+16?

- hartnn

no ? how u got that ?

- anonymous

multiplied...

- hartnn

derivative of x^3+3x+8
lets take it term by term
derivative of x^3 =... ?

- anonymous

3x

- hartnn

no...
derivative of x^n = nx^{n-1}

- anonymous

3x^2

- hartnn

derivative of 3x ?

- anonymous

3

- hartnn

derivative of 8 ?

- anonymous

2? i really am so confused

- hartnn

derivative of a constant = 0
so derivative of 8 is 0
so, derivative of x^3+3x+8 = 3x^2+3
right ?

- anonymous

oh okay.

- anonymous

i understand that better.

- anonymous

now what if 2 was plugged in

- hartnn

so, we have :
derivative of (x^3+3x+8)^2 will be 2(x^3+3x+8) *(derivative of x^3+3x+8 ) ,
= 2(x^3+3x+8) * (3x^2+3)
got this ?

- anonymous

i thought the derivative of f '(x) is 3x^2+3

- hartnn

didn't i mention to u the chain rule ?

- anonymous

oh yeah, multiple use the one before and after

- hartnn

so any doubts now ? that would be your final answer...

- anonymous

nope no doubts. now what if the problem was f ' (2)= ?

- hartnn

just plug in x=2

- anonymous

i did but im still not sure what am i suppose to do.

- anonymous

i plug them in but i get confused with the derivative and actually multiplying it out.

- hartnn

2(x^3+3x+8) * (3x^2+3)
put x=2
2(2^3+3*2+8) * (3*2^2+3) =... ?

- anonymous

44

- hartnn

i get 660 :P
using google
https://www.google.co.in/search?q=2(2%5E3%2B3*2%2B8)+*+(3*2%5E2%2B3)&aq=f&oq=2(2%5E3%2B3*2%2B8)+*+(3*2%5E2%2B3)&aqs=chrome.0.57.192&sourceid=chrome&ie=UTF-8

- anonymous

thank you. thats correct

- hartnn

welcome ^_^

- anonymous

would it be possible to know the answer of a few problems because I would need to sign off and practice by hand but I would like to check my work and see if they're correct.

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