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

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do you know the chain rule ?
not really sure with the chain rule..
need to learn that then to do a question of this type

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Other answers:

yes, that's why I'm trying to learn it here.. if that helps.
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 ?
by definition, \(f(g(x))' = f'(g(x)) \times g'(x)\)
so try once for your Q, maybe.. ?
you dont subtract the exponent n-1?
i made the exponent n-1 in the Q, ax^2+bx+c had exponent 3, in the answer i became 2, right ?
for x^n, the derivative is \((x^n)'=n x^{n-1}\)
okay... sort of there. but then the second equation next to it confused me a little.
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
in your Q, you'll multiply by (x^3+3x+8)
**by the derivative of (x^3+3x+8)
so the derivative for F'(x) would be.. (3x^2+3x+8)?
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
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 ?
no ? how u got that ?
derivative of x^3+3x+8 lets take it term by term derivative of x^3 =... ?
no... derivative of x^n = nx^{n-1}
derivative of 3x ?
derivative of 8 ?
2? i really am so confused
derivative of a constant = 0 so derivative of 8 is 0 so, derivative of x^3+3x+8 = 3x^2+3 right ?
oh okay.
i understand that better.
now what if 2 was plugged in
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 ?
i thought the derivative of f '(x) is 3x^2+3
didn't i mention to u the chain rule ?
oh yeah, multiple use the one before and after
so any doubts now ? that would be your final answer...
nope no doubts. now what if the problem was f ' (2)= ?
just plug in x=2
i did but im still not sure what am i suppose to do.
i plug them in but i get confused with the derivative and actually multiplying it out.
2(x^3+3x+8) * (3x^2+3) put x=2 2(2^3+3*2+8) * (3*2^2+3) =... ?
i get 660 :P using google*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
thank you. thats correct
welcome ^_^
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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