mathcalculus
can someone help me solve this problem ?
let f(x)= (x^3+3x+8)^2
F'(x)=
F'(2)=
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hartnn
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do you know the chain rule ?
mathcalculus
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not really sure with the chain rule..
sam1194
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need to learn that then to do a question of this type
mathcalculus
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yes, that's why I'm trying to learn it here.. if that helps.
hartnn
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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
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by definition,
\(f(g(x))' = f'(g(x)) \times g'(x)\)
mathcalculus
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yeah
hartnn
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so try once for your Q, maybe.. ?
mathcalculus
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you dont subtract the exponent n-1?
hartnn
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i made the exponent n-1
in the Q, ax^2+bx+c had exponent 3, in the answer i became 2, right ?
hartnn
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for x^n, the derivative is
\((x^n)'=n x^{n-1}\)
mathcalculus
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okay... sort of there. but then the second equation next to it confused me a little.
hartnn
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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
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in your Q, you'll multiply by (x^3+3x+8)
hartnn
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**by the derivative of (x^3+3x+8)
mathcalculus
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so the derivative for F'(x) would be.. (3x^2+3x+8)?
hartnn
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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
mathcalculus
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2x
hartnn
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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 ?
mathcalculus
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2x^2+6x+16?
hartnn
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no ? how u got that ?
mathcalculus
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multiplied...
hartnn
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derivative of x^3+3x+8
lets take it term by term
derivative of x^3 =... ?
mathcalculus
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3x
hartnn
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no...
derivative of x^n = nx^{n-1}
mathcalculus
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3x^2
hartnn
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derivative of 3x ?
mathcalculus
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3
hartnn
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derivative of 8 ?
mathcalculus
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2? i really am so confused
hartnn
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derivative of a constant = 0
so derivative of 8 is 0
so, derivative of x^3+3x+8 = 3x^2+3
right ?
mathcalculus
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oh okay.
mathcalculus
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i understand that better.
mathcalculus
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now what if 2 was plugged in
hartnn
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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 ?
mathcalculus
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i thought the derivative of f '(x) is 3x^2+3
hartnn
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didn't i mention to u the chain rule ?
mathcalculus
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oh yeah, multiple use the one before and after
hartnn
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so any doubts now ? that would be your final answer...
mathcalculus
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nope no doubts. now what if the problem was f ' (2)= ?
hartnn
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just plug in x=2
mathcalculus
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i did but im still not sure what am i suppose to do.
mathcalculus
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i plug them in but i get confused with the derivative and actually multiplying it out.
hartnn
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2(x^3+3x+8) * (3x^2+3)
put x=2
2(2^3+3*2+8) * (3*2^2+3) =... ?
mathcalculus
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44
mathcalculus
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thank you. thats correct
hartnn
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welcome ^_^
mathcalculus
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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.