Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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, \(f(g(x))' = f'(g(x)) \times g'(x)\)
yeah
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
2x
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 ?
2x^2+6x+16?
no ? how u got that ?
multiplied...
derivative of x^3+3x+8 lets take it term by term derivative of x^3 =... ?
3x
no... derivative of x^n = nx^{n-1}
3x^2
derivative of 3x ?
3
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) =... ?
44
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
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question