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anonymous

  • 5 years ago

Evaluate: anti-deriviative 3/(x-4) dx will I use U substitute

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  1. anonymous
    • 5 years ago
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    Yes.

  2. anonymous
    • 5 years ago
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    Identify your composite functions f(a) and g(x) such that 3/(x-4) = f(g(x))

  3. anonymous
    • 5 years ago
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    u= x-4 du= 1

  4. anonymous
    • 5 years ago
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    Yes.

  5. anonymous
    • 5 years ago
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    Well, actually not quite

  6. anonymous
    • 5 years ago
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    du = 1dx

  7. anonymous
    • 5 years ago
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    yes I had that just forgot to type it

  8. anonymous
    • 5 years ago
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    ok good.

  9. anonymous
    • 5 years ago
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    ok this is where I have trouble.....will it be 3/u ???? Doesn't look right

  10. anonymous
    • 5 years ago
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    That is exactly right. If u = x-4 then 3/(x-4) = 3/u

  11. anonymous
    • 5 years ago
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    so will I add 1 to the u (exponent) to be u^2

  12. anonymous
    • 5 years ago
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    Not quite. realize that you basically have \[3 * \frac{1}{u}\] So the derivative of what function of u gives you \(\frac{1}{u}\)?

  13. anonymous
    • 5 years ago
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    u^-1 ??

  14. anonymous
    • 5 years ago
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    1/u is u^-1

  15. anonymous
    • 5 years ago
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    So it's not that. \(\frac{d}{du}[ln\ u] = \frac{1}{u}\) remember?

  16. anonymous
    • 5 years ago
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    not sure what you mean about that?

  17. anonymous
    • 5 years ago
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    The derivative of ln x = 1/x right?

  18. anonymous
    • 5 years ago
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    ln is the natural logarithm function.

  19. anonymous
    • 5 years ago
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    oh, yes it does. Too many rules.....

  20. anonymous
    • 5 years ago
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    yeah, you'll get them with practice.

  21. anonymous
    • 5 years ago
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    So the anti derivative of 1/x = ln x + C

  22. anonymous
    • 5 years ago
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    So 3 times the anti derivative of 1/u du = ?

  23. anonymous
    • 5 years ago
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    ok lost again..... I know we have 3/u or 3 * 1/u would it be 3lnx + c ???

  24. anonymous
    • 5 years ago
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    It would be 3(ln u) + C, but then you have to plug in what u is.

  25. anonymous
    • 5 years ago
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    how about 3 ln\[\left| x-4 \right|\] + C sorry couln't get them all in one line

  26. anonymous
    • 5 years ago
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    Yep, that's it exactly.

  27. anonymous
    • 5 years ago
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    I am getting there......slowly....... Thanks I really appreciate your help.

  28. anonymous
    • 5 years ago
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    Of course! Just keep practicing. You're doing very well

  29. anonymous
    • 5 years ago
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    I have a test next Tue. So I will be asking questions up until then. I am sure I will talk to you again..

  30. anonymous
    • 5 years ago
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    ok!

  31. anonymous
    • 5 years ago
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    I got really good with the derivatives, so my brain still wants to do them.

  32. anonymous
    • 5 years ago
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    Yeah, I know the feeling. Soon you'll get the hang of going the other way.

  33. anonymous
    • 5 years ago
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    I will start one more question before I have to leave

  34. anonymous
    • 5 years ago
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    ok

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