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kobe24

  • 3 years ago

what is the derivative of ln(1+1/x)? they say the answer is xsquared-1/ xcubed+x now can someone explain how you find the answer

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  1. dpaInc
    • 3 years ago
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    the answer given is correct but not simplified.... use the formula \(\large [lny]'=\frac{y'}{y} \) to find the derivative...

  2. kobe24
    • 3 years ago
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    then how do i use this formula for the problem

  3. dpaInc
    • 3 years ago
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    let \(\large y=1+\frac{1}{x} \) so, \(\large [ln(1+\frac{1}{x})]'=\frac{[1+\frac{1}{x}]'}{1+\frac{1}{x}} \) can you find \(\large [1+\frac{1}{x}]'= \) ????

  4. kobe24
    • 3 years ago
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    no how do u find 1+1/x? is the answer -1x-1/2

  5. dpaInc
    • 3 years ago
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    use the sum rule for derivatives... \(\large [1+\frac{1}{x}]'=[1]'+[\frac{1}{x}]' \) continue....

  6. kobe24
    • 3 years ago
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    |dw:1357167466355:dw| ??

  7. dpaInc
    • 3 years ago
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    ok... good... so, \(\large [ln(1+\frac{1}{x})]'=\frac{[1+\frac{1}{x}]'}{1+\frac{1}{x}}=\frac{-x^{(\frac{-1}{2})}}{1+\frac{1}{x}} \) now just simplify...

  8. kobe24
    • 3 years ago
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    wait but how do i simplify to get ...|dw:1357167740926:dw|

  9. dpaInc
    • 3 years ago
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    factor the top and bottom...

  10. PhoenixFire
    • 3 years ago
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    \[[{x^n}]^{\prime }=nx^{n-1}\] How did you get\[-x^{-{1\over 2}}\] shouldn't it be \[-x^{-2}=-{1\over x^2}\]

  11. dpaInc
    • 3 years ago
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    oops.... sorry.... @PhoenixFire is correct...

  12. kobe24
    • 3 years ago
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    so what should i do should it look like this...|dw:1357167900420:dw|

  13. dpaInc
    • 3 years ago
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    yes.. ^^^, now simplify....

  14. kobe24
    • 3 years ago
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    how....|dw:1357167986477:dw|

  15. dpaInc
    • 3 years ago
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    try rewriting the denominator as...|dw:1357168045359:dw|

  16. kobe24
    • 3 years ago
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    then can i just write the answer....|dw:1357168177116:dw| or no?!?!

  17. kobe24
    • 3 years ago
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    how do i continue simplifying from what i have now?

  18. dpaInc
    • 3 years ago
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    i'm double checking.... looks like this is wrong.... |dw:1357168325298:dw|

  19. kobe24
    • 3 years ago
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    well it says that is the derivative and answer and its never been wrong

  20. PhoenixFire
    • 3 years ago
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    Every time I do it I get \(\large -{1 \over x^2+x}\) I think the answer that it's given you is wrong. Unless you are accidentally looking at the wrong question in the answer section. I do that all the time.

  21. dpaInc
    • 3 years ago
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    the answer is wrong.... if you continue what we have, you'd get the answer produced by @PhoenixFire ...

  22. dpaInc
    • 3 years ago
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    and wolfram... http://www.wolframalpha.com/input/?i=derivative+ln%281%2B1%2Fx%29

  23. dpaInc
    • 3 years ago
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    just continue what we have here... |dw:1357168741789:dw|

  24. kobe24
    • 3 years ago
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    yeah i believe both of u now. okay ill just tell my teacher that it is wrong.

  25. kobe24
    • 3 years ago
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    so from the pic that u just put up

  26. dpaInc
    • 3 years ago
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    yes... just continue that and you'll get the correct answer..

  27. kobe24
    • 3 years ago
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    idk exactly how to though

  28. kobe24
    • 3 years ago
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    okay i was talking about the ln not log though to start the problem... is that original answer wrong still?

  29. dpaInc
    • 3 years ago
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    \(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{x^2}\cdot \frac{x}{x+1} \)

  30. kobe24
    • 3 years ago
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    |dw:1357169117534:dw|

  31. kobe24
    • 3 years ago
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    thanks for the help

  32. dpaInc
    • 3 years ago
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    \(\huge \frac{\frac{-1}{x^2}}{\frac{x+1}{x}}=\frac{-1}{\cancel {x^2}x}\cdot \frac{\cancel x1}{x+1} \)

  33. dpaInc
    • 3 years ago
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    @PhoenixFire , thanks for getting my back.... :) i'm a little rusty in calc... :(

  34. malevolence19
    • 3 years ago
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    I like the complete lack of trying and still no understanding from this question.

  35. Rowa
    • 3 years ago
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    this was funny to watch

  36. malevolence19
    • 3 years ago
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    When people don't try they don't succeed. It's that simple.

  37. Rowa
    • 3 years ago
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    still funny seeing people trying to explain to someone thats lazy

  38. GCR92
    • 3 years ago
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    I don't know if your teacher want's you to simplify that, that's not the kind of "simplifying exercise" it looks like it's just to learn the Ln derivative, I took a picture, sorry i'm very traditional, hope it helps you.

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  39. GCR92
    • 3 years ago
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    There you go, I hope it's clear now

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