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jaersyn

  • 3 years ago

What is the derivative of: (secu)^(n-2)? Is it (n-2)(secu)^(n-3) tanu?

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  1. jaersyn
    • 3 years ago
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    n is an integer

  2. hsmt
    • 3 years ago
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    hi. oh, ok

  3. hsmt
    • 3 years ago
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    what about u?

  4. jaersyn
    • 3 years ago
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    just a variable

  5. hsmt
    • 3 years ago
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    not a function of x?

  6. hsmt
    • 3 years ago
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    ok, u is not a function of x, then i think you are gonna want to check the exponent of sec u

  7. hsmt
    • 3 years ago
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    still there?

  8. jaersyn
    • 3 years ago
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    yeah

  9. jaersyn
    • 3 years ago
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    i got it

  10. hsmt
    • 3 years ago
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    in case you get this, everything looks ok except the exponent on sec u. the n-2 out front is good, the sec^(n-3) times the sec u tan u (chain rule) is what is affecting the exponent

  11. hsmt
    • 3 years ago
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    do you see what the exponent is now for sec u?

  12. jaersyn
    • 3 years ago
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    it's n-3 no?

  13. hsmt
    • 3 years ago
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    the n-2 in front is correct. you also have product of secu^(n-3) and then there is the chain rule that multiplies by secutanu. so, it is not n-3

  14. hsmt
    • 3 years ago
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    do you see what i am trying to point out? feel free to ask any questions

  15. hsmt
    • 3 years ago
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    are you still having difficulty?

  16. jaersyn
    • 3 years ago
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    oh pellet

  17. jaersyn
    • 3 years ago
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    you are correct

  18. jaersyn
    • 3 years ago
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    it would still be sec^(n-2)

  19. hsmt
    • 3 years ago
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    niiice

  20. jaersyn
    • 3 years ago
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    (n-2)(secu)^(n-3) times secutanu which equals (n-2)(secu)^(n-2) times tanu

  21. hsmt
    • 3 years ago
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    very cool you got it

  22. jaersyn
    • 3 years ago
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    now i have to rewrite this wholeeeeeeeeeee thing......... thanks

  23. hsmt
    • 3 years ago
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    dontcha just hate the whole rewriting thing? If there is nothing else i can help with, i will bid you adieu

  24. hsmt
    • 3 years ago
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    take care

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