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calyne Group Title

Differentiate the function f(x) = (lnx)/(1+ln(2x))

  • 2 years ago
  • 2 years ago

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  1. Sarkar Group Title
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    use quotient rule...

    • 2 years ago
  2. bmp Group Title
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    1/(x(lnx+1)^2) by using the quotient rule :-)

    • 2 years ago
  3. bmp Group Title
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    Oops, sorry

    • 2 years ago
  4. calyne Group Title
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    i'm trying it i'm not sure if i'm getting it just show me so i can check

    • 2 years ago
  5. Sarkar Group Title
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    |dw:1334417837384:dw|

    • 2 years ago
  6. calyne Group Title
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    like that helps

    • 2 years ago
  7. bmp Group Title
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    It's (1+ln2)/(x(ln2x + 1)^2)

    • 2 years ago
  8. calyne Group Title
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    how did you get that

    • 2 years ago
  9. Sarkar Group Title
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    |dw:1334418069246:dw|

    • 2 years ago
  10. calyne Group Title
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    so then [(1+ln(x)) / x] / [(1+ln(2x))^2] is what

    • 2 years ago
  11. calyne Group Title
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    i mean (1+ln(2)) / x

    • 2 years ago
  12. bmp Group Title
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    http://www.wolframalpha.com/input/?i=d%28lnx%2F%281%2Bln2x%29%29%2Fdx Take a look at show steps :-)

    • 2 years ago
  13. Sarkar Group Title
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    ([(1+ln(2x)) / x]-[(lnx)/x] )/ [(1+ln(2x))^2]

    • 2 years ago
  14. TuringTest Group Title
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    super secret tip: the quotient rule is best avoided. better to rewrite the problem and use the product rule

    • 2 years ago
  15. Sarkar Group Title
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    but that involves confusion using the power of -1 ,i feel...

    • 2 years ago
  16. TuringTest Group Title
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    yes, so if you know the chain rule this shouldn't be too hard\[\ln x(1+\ln(2x))^{-1}\]

    • 2 years ago
  17. Sarkar Group Title
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    thanks for informin@Turingtest.

    • 2 years ago
  18. TuringTest Group Title
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    so @calyne try this with product rule\[\ln x(1+\ln(2x))^{-1}\]let\[u=\ln x\]\[v=(1+\ln(2x))^{-1}\]

    • 2 years ago
  19. calyne Group Title
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    oh nvm nvm sry oh god

    • 2 years ago
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