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anonymous

  • 5 years ago

Differentiate the given function: f(x)= ln (x+1)/(x-1) I get -2/(x-1)^2

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  1. anonymous
    • 5 years ago
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    -2/ (x^2-1)

  2. anonymous
    • 5 years ago
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    I used the quotient rule, but not sure what to do with the ln

  3. anonymous
    • 5 years ago
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    did u have? ln(x+1/x-1)?

  4. anonymous
    • 5 years ago
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    if yes, then d/dx (lnx)= 1/x ... easy

  5. anonymous
    • 5 years ago
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    the ln is before the fraction

  6. anonymous
    • 5 years ago
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    Quotient rule u= ln (x+1) v= ln (x-1)

  7. anonymous
    • 5 years ago
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    If ln is over the whole thing ln [(x+1)/(x-1)] rewrite as [ln (x+1)]/[ln (x-1)]

  8. anonymous
    • 5 years ago
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    I don't know if this will help ln x+1 x-1 x+1/x-1 one big parenthesis around this fraction

  9. anonymous
    • 5 years ago
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    Can be rewritten as \[\ln (x+1)\div \ln (x-1)\]

  10. anonymous
    • 5 years ago
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    Top is u, bottom v

  11. anonymous
    • 5 years ago
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    I have not done it that way

  12. anonymous
    • 5 years ago
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    the u and v

  13. anonymous
    • 5 years ago
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    what shorthand you use for quotient rule f(x) or what?

  14. anonymous
    • 5 years ago
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    ok for quotient rule, I can do that but not sure what to do with the ln. After the quotient rule I get (x-1)-(x+1) / (x-1)^2

  15. anonymous
    • 5 years ago
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    Excuse me for using u and v. Let u=ln (x+1), let v=ln (x-1) u'=?, v'=? u' means u prime or derivative of u The derivative of the whole thing is \[(u'v-uv')\div(v ^{2})\]

  16. anonymous
    • 5 years ago
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    I am sorry but I am having a real hard time understanding this math. I have a book that has this problem worked out. I just don't understand why certain things are done the way they are. The last section we went over was dervatives and I did really well with them. These I am not.

  17. anonymous
    • 5 years ago
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    OK, so write one or two steps from book and say what you don't understand

  18. anonymous
    • 5 years ago
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    1st step f'(x) = 1/(x+1/x-1) d/dx (x+1/x-1) why did they put 1 over the problem.

  19. anonymous
    • 5 years ago
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    next step x-1/x+1 [(x-1)(1) - (x+1)(1)//(x-1)^2 the 2nd half I get, that is the quotient rule

  20. anonymous
    • 5 years ago
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    they are treating the whole fraction thingy as a single number. Let (x+1)/(x-1) be u (sorry) f' of ln u=1/u

  21. anonymous
    • 5 years ago
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    Deal with one question at a time or everything would get confused

  22. anonymous
    • 5 years ago
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    ok with step one 1/(x+1/x-1) then I guess they took the bottom # and flipped and multiplied. Is that how the problem got switched for step 2

  23. anonymous
    • 5 years ago
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    OK, back to original question what might make it a little confusing is what is called chain rule. the f' (x) = 1/[x+1/(x-1)] du in addition to this you must find the derivative of the inner function (x+1)/(x-1). Find the derivative of that and that is your du. This is a complicated function. I can explain more after you look at it.

  24. anonymous
    • 5 years ago
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    why can't I use the quotient rule

  25. anonymous
    • 5 years ago
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    by putting the 1 / .... is that taking the ln out of the problem

  26. anonymous
    • 5 years ago
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    There is a lot going on in this problem. It is not that you can't use quotient rule. In fact they are using quotient rule. But there are some intricacies that make their answer difficult to understand. Start from the top. Ask one piece by one piece what you don't understand.

  27. anonymous
    • 5 years ago
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    ok 1st step 1/ (x+1/x-1) if I use the quotient rule derv. is x+1/x-1

  28. anonymous
    • 5 years ago
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    never mind that isn't correct.

  29. anonymous
    • 5 years ago
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    Just write the book line by line and I will do the play by play

  30. anonymous
    • 5 years ago
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    1st step f'(x) = 1/(x+1/x-1) d/dx (x+1/x-1) 2nd step x-1/x+1 [(x-1)(1) - (x+1)(1)/(x-1)^2 step 3 -2/(x+1)(x-1) That is all steps

  31. anonymous
    • 5 years ago
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    Are you familiar with the thing called the chain rule?

  32. anonymous
    • 5 years ago
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    yes

  33. anonymous
    • 5 years ago
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    There approach is good. Now that I see it. They considered the ln (that thing) ln (that thing)=[1/(that thing)][multiplied by derivative of that thing] They didn't find it necessary to do the quotient rule to find f'(x), but in finding the derivative of (that thing) they did the quotient rule. Derivative of ln doesn't require work, you just put 1/something

  34. anonymous
    • 5 years ago
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    ok I half way understand that but how did they get x-1/x+1 in step 2 before the quotient rule

  35. anonymous
    • 5 years ago
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    sometimes when they work the problems out in the book it doesn't explain what they are doing. If I understand the work it does not confuss me.

  36. anonymous
    • 5 years ago
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    1/(B/C)=C/B

  37. anonymous
    • 5 years ago
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    ?

  38. anonymous
    • 5 years ago
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    You have fractional property at the bottom is flipped.

  39. anonymous
    • 5 years ago
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    ok you flip it to get rid of the 1

  40. anonymous
    • 5 years ago
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    yes

  41. anonymous
    • 5 years ago
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    ok on step 2 x-1/x+1 [(x-1)(1)-(x+1)(1)/(x-1)^2 ok I can see the quotient rule what are they doine with the x-1/x+1, are they multipling that by the quotient?

  42. anonymous
    • 5 years ago
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    never mind something just clicked, I see it now. They cross multiplied

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