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anonymous

  • 5 years ago

Hey guys I'm having problems with derivatives if someone could explain the logx/lnx derivatives to me that would be great heres a few examples y = -2x^2(lnx)^4 or y = ln((x +1)/x)

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  1. anonymous
    • 5 years ago
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    to start, do you know the derivative of \[\ln(x)\] ?

  2. anonymous
    • 5 years ago
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    \[d/dx [ \ln (x) ] = 1/x\]

  3. anonymous
    • 5 years ago
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    yes i just dont understand how it applies when you have things attached to x

  4. anonymous
    • 5 years ago
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    such as the (lnx)^4 or the ln(x +1/x)

  5. anonymous
    • 5 years ago
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    ok

  6. anonymous
    • 5 years ago
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    let's try \[\ln((x+1)/x)\]

  7. anonymous
    • 5 years ago
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    alright

  8. anonymous
    • 5 years ago
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    have you used substitution rule before?

  9. anonymous
    • 5 years ago
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    That example is very simple, though the process might be tedious. Simply put a 1 over the expression then multiply by derivative of the expression

  10. anonymous
    • 5 years ago
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    No, i got that wrong, i was describing log.

  11. anonymous
    • 5 years ago
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    I'm contradicting myself, I think i told you right

  12. anonymous
    • 5 years ago
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    so, ln((x+1)/x)

  13. anonymous
    • 5 years ago
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    using the substitution rule: let's have \[u = (x+1)/x , du = -1/x^2\]

  14. anonymous
    • 5 years ago
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    du = =1/x^2 * dx

  15. anonymous
    • 5 years ago
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    yes, addmaster?

  16. anonymous
    • 5 years ago
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    ok sorry the substitution thing is confusing me

  17. anonymous
    • 5 years ago
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    ok, well, the concept is thus: essentially you can simply things by taking a difficult expression to immeditaely differentitate, like here ln((x+1)/x), and simplify the first steps a bit

  18. anonymous
    • 5 years ago
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    outright, [ln((x+1)/x)]' looks difficult, but [ln(u)]'

  19. anonymous
    • 5 years ago
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    that is like a calc identity

  20. anonymous
    • 5 years ago
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    ok i get it

  21. anonymous
    • 5 years ago
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    the trick is keeping track of this substitution, you'll need to find the derivtative of u. as above, u=(x+1)/x, while du = (-1/x^2)*dx

  22. anonymous
    • 5 years ago
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    ahhh ok so if you set it as a variable it complicates things less

  23. anonymous
    • 5 years ago
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    so we have our first equation: ln((x+1)/x) dx

  24. anonymous
    • 5 years ago
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    yeah, you do less internal variable manipulation

  25. anonymous
    • 5 years ago
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    alright got it sorry

  26. anonymous
    • 5 years ago
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    cool, you got it?

  27. anonymous
    • 5 years ago
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    yep that helped a lottt

  28. anonymous
    • 5 years ago
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    so, you want to solve it? i can stay to help if you need it...

  29. anonymous
    • 5 years ago
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    but just to be sure once I have the x + 1 over x I have to then use quotient rule...?

  30. anonymous
    • 5 years ago
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    so, u = (x+1)/x ... but that equals x/x + 1/x

  31. anonymous
    • 5 years ago
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    so u = 1 + 1/x

  32. anonymous
    • 5 years ago
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    du = (1/x) dx

  33. anonymous
    • 5 years ago
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    er, (1/x)'

  34. anonymous
    • 5 years ago
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    which is (\[x^-1\])'

  35. anonymous
    • 5 years ago
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    so, whats the derivative with respect to x of 1 + x^-1 ?

  36. anonymous
    • 5 years ago
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    -1x^-2...?

  37. anonymous
    • 5 years ago
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    yeah

  38. anonymous
    • 5 years ago
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    so, ln(u) du ---> but du = (-1/x^2)*dx

  39. anonymous
    • 5 years ago
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    ln(u)' --> 1/u, where u is (1 + 1/x) ....so you have \[1/[(1 + 1/x)]\]....then, you can't forget that du was -1/x^2 * dx, so when you now substitute back in for u so you have the original variable x, you have to multiply by (-1/x^2)

  40. anonymous
    • 5 years ago
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    \[\frac{1}{(1+1/x)} * \frac{-1}{(x^2)}\]

  41. anonymous
    • 5 years ago
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    which you can simplify

  42. anonymous
    • 5 years ago
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    to : -1/(x^2 + x)

  43. anonymous
    • 5 years ago
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    ok sorry my computers really crappy so i cant reply fast but i completely get it now thanks!

  44. anonymous
    • 5 years ago
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    no problem, hopefully now you can use the book examples much better too!

  45. anonymous
    • 5 years ago
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    also, here on OpenStudy, if someone gives you good help, you can award them with a medal here in the conversation

  46. anonymous
    • 5 years ago
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    so, ask more questions and maybe you can help some other math learners if you see a question you know to answer

  47. anonymous
    • 5 years ago
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    i'm off, but good luck with the calc ;)

  48. anonymous
    • 5 years ago
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    thank you very much

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