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anonymous

  • 4 years ago

Let F(x)=f(x^ 3 ) and G(x)=(f(x)) ^3 . You also know that a ^2 =4,f(a)=2,f ′ (a)=2,f ′ (a ^3 )=6 . Find F'(a) and G'(a)

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  1. TuringTest
    • 4 years ago
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    do you have any idea how I got the formula u'(x)=f'(g(x))g'(x) in the last problem?

  2. anonymous
    • 4 years ago
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    no

  3. TuringTest
    • 4 years ago
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    that is the chain rule I don't see how you can solve these problems without it: \[\large [f(g(x))]'=f'(g(x))g'(x)\]

  4. TuringTest
    • 4 years ago
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    we differentiate our way inward when we have nested functions like this -derivative of the outer function times the derivative of the inner function

  5. anonymous
    • 4 years ago
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    ok

  6. TuringTest
    • 4 years ago
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    so see if you can apply that to find the derivative of\[\large F(x)=f(x^3)\]

  7. anonymous
    • 4 years ago
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    3x^2

  8. TuringTest
    • 4 years ago
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    not quite, it's 3x^2f'(x^3) by the chain rule

  9. anonymous
    • 4 years ago
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    oh ok

  10. TuringTest
    • 4 years ago
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    we start with the derivative of the outer function, then the inner one so F'(a)=3a^2f'(a^3) and we are given all those quantities in the problem, so plug in and solve

  11. anonymous
    • 4 years ago
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    3(4)(6)

  12. anonymous
    • 4 years ago
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    72

  13. TuringTest
    • 4 years ago
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    yes!

  14. TuringTest
    • 4 years ago
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    now try to come up with an expression for G'(x) remember the chain rule

  15. anonymous
    • 4 years ago
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    3f(x)^2f'(x)

  16. TuringTest
    • 4 years ago
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    nice! now you've got those quantities as well, so what's G'(a) ?

  17. anonymous
    • 4 years ago
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    24

  18. anonymous
    • 4 years ago
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    thanks for all of the help. i really appreciate your patience with me, especially the last problem where i was making alot of mistakes

  19. anonymous
    • 4 years ago
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    alot of the people whom i seek help from would just belittle me

  20. anonymous
    • 4 years ago
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    it has discourage me from seeking help face to face

  21. TuringTest
    • 4 years ago
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    you have done most of the work yourself, don't let them get you down you picked that up at lightning speed :D

  22. TuringTest
    • 4 years ago
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    new things are always tricky, I'm just glad I could help good luck, and don't forget the chain rule!

  23. anonymous
    • 4 years ago
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    alright thanks. i am done for the day. take care

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