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mathcalculus

  • 3 years ago

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

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  1. hartnn
    • 3 years ago
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    do you know the chain rule ?

  2. mathcalculus
    • 3 years ago
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    not really sure with the chain rule..

  3. sam1194
    • 3 years ago
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    need to learn that then to do a question of this type

  4. mathcalculus
    • 3 years ago
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    yes, that's why I'm trying to learn it here.. if that helps.

  5. hartnn
    • 3 years ago
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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 ?

  6. hartnn
    • 3 years ago
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    by definition, \(f(g(x))' = f'(g(x)) \times g'(x)\)

  7. mathcalculus
    • 3 years ago
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    yeah

  8. hartnn
    • 3 years ago
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    so try once for your Q, maybe.. ?

  9. mathcalculus
    • 3 years ago
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    you dont subtract the exponent n-1?

  10. hartnn
    • 3 years ago
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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 ?

  11. hartnn
    • 3 years ago
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    for x^n, the derivative is \((x^n)'=n x^{n-1}\)

  12. mathcalculus
    • 3 years ago
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    okay... sort of there. but then the second equation next to it confused me a little.

  13. hartnn
    • 3 years ago
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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

  14. hartnn
    • 3 years ago
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    in your Q, you'll multiply by (x^3+3x+8)

  15. hartnn
    • 3 years ago
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    **by the derivative of (x^3+3x+8)

  16. mathcalculus
    • 3 years ago
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    so the derivative for F'(x) would be.. (3x^2+3x+8)?

  17. hartnn
    • 3 years ago
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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

  18. mathcalculus
    • 3 years ago
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    2x

  19. hartnn
    • 3 years ago
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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 ?

  20. mathcalculus
    • 3 years ago
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    2x^2+6x+16?

  21. hartnn
    • 3 years ago
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    no ? how u got that ?

  22. mathcalculus
    • 3 years ago
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    multiplied...

  23. hartnn
    • 3 years ago
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    derivative of x^3+3x+8 lets take it term by term derivative of x^3 =... ?

  24. mathcalculus
    • 3 years ago
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    3x

  25. hartnn
    • 3 years ago
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    no... derivative of x^n = nx^{n-1}

  26. mathcalculus
    • 3 years ago
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    3x^2

  27. hartnn
    • 3 years ago
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    derivative of 3x ?

  28. mathcalculus
    • 3 years ago
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    3

  29. hartnn
    • 3 years ago
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    derivative of 8 ?

  30. mathcalculus
    • 3 years ago
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    2? i really am so confused

  31. hartnn
    • 3 years ago
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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 ?

  32. mathcalculus
    • 3 years ago
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    oh okay.

  33. mathcalculus
    • 3 years ago
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    i understand that better.

  34. mathcalculus
    • 3 years ago
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    now what if 2 was plugged in

  35. hartnn
    • 3 years ago
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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 ?

  36. mathcalculus
    • 3 years ago
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    i thought the derivative of f '(x) is 3x^2+3

  37. hartnn
    • 3 years ago
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    didn't i mention to u the chain rule ?

  38. mathcalculus
    • 3 years ago
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    oh yeah, multiple use the one before and after

  39. hartnn
    • 3 years ago
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    so any doubts now ? that would be your final answer...

  40. mathcalculus
    • 3 years ago
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    nope no doubts. now what if the problem was f ' (2)= ?

  41. hartnn
    • 3 years ago
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    just plug in x=2

  42. mathcalculus
    • 3 years ago
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    i did but im still not sure what am i suppose to do.

  43. mathcalculus
    • 3 years ago
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    i plug them in but i get confused with the derivative and actually multiplying it out.

  44. hartnn
    • 3 years ago
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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) =... ?

  45. mathcalculus
    • 3 years ago
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    44

  46. mathcalculus
    • 3 years ago
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    thank you. thats correct

  47. hartnn
    • 3 years ago
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    welcome ^_^

  48. mathcalculus
    • 3 years ago
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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.

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