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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

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

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.

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)?

2x

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 ?

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

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

thank you. thats correct

welcome ^_^