anonymous
  • anonymous
can someone help me solve this problem ? let f(x)= (x^3+3x+8)^2 F'(x)= F'(2)=
Calculus1
chestercat
  • chestercat
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hartnn
  • hartnn
do you know the chain rule ?
anonymous
  • anonymous
not really sure with the chain rule..
anonymous
  • anonymous
need to learn that then to do a question of this type

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anonymous
  • anonymous
yes, that's why I'm trying to learn it here.. if that helps.
hartnn
  • 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
  • hartnn
by definition, \(f(g(x))' = f'(g(x)) \times g'(x)\)
anonymous
  • anonymous
yeah
hartnn
  • hartnn
so try once for your Q, maybe.. ?
anonymous
  • anonymous
you dont subtract the exponent n-1?
hartnn
  • 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
  • hartnn
for x^n, the derivative is \((x^n)'=n x^{n-1}\)
anonymous
  • anonymous
okay... sort of there. but then the second equation next to it confused me a little.
hartnn
  • 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
  • hartnn
in your Q, you'll multiply by (x^3+3x+8)
hartnn
  • hartnn
**by the derivative of (x^3+3x+8)
anonymous
  • anonymous
so the derivative for F'(x) would be.. (3x^2+3x+8)?
hartnn
  • 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
  • anonymous
2x
hartnn
  • 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
  • anonymous
2x^2+6x+16?
hartnn
  • hartnn
no ? how u got that ?
anonymous
  • anonymous
multiplied...
hartnn
  • hartnn
derivative of x^3+3x+8 lets take it term by term derivative of x^3 =... ?
anonymous
  • anonymous
3x
hartnn
  • hartnn
no... derivative of x^n = nx^{n-1}
anonymous
  • anonymous
3x^2
hartnn
  • hartnn
derivative of 3x ?
anonymous
  • anonymous
3
hartnn
  • hartnn
derivative of 8 ?
anonymous
  • anonymous
2? i really am so confused
hartnn
  • hartnn
derivative of a constant = 0 so derivative of 8 is 0 so, derivative of x^3+3x+8 = 3x^2+3 right ?
anonymous
  • anonymous
oh okay.
anonymous
  • anonymous
i understand that better.
anonymous
  • anonymous
now what if 2 was plugged in
hartnn
  • 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
  • anonymous
i thought the derivative of f '(x) is 3x^2+3
hartnn
  • hartnn
didn't i mention to u the chain rule ?
anonymous
  • anonymous
oh yeah, multiple use the one before and after
hartnn
  • hartnn
so any doubts now ? that would be your final answer...
anonymous
  • anonymous
nope no doubts. now what if the problem was f ' (2)= ?
hartnn
  • hartnn
just plug in x=2
anonymous
  • anonymous
i did but im still not sure what am i suppose to do.
anonymous
  • anonymous
i plug them in but i get confused with the derivative and actually multiplying it out.
hartnn
  • hartnn
2(x^3+3x+8) * (3x^2+3) put x=2 2(2^3+3*2+8) * (3*2^2+3) =... ?
anonymous
  • anonymous
44
hartnn
  • 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
  • anonymous
thank you. thats correct
hartnn
  • hartnn
welcome ^_^
anonymous
  • 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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