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calyne

  • 2 years ago

Differentiate f and find the domain of f. f(x) = x/[1-ln(x-1)]

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  1. experimentX
    • 2 years ago
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    use quotient rule.

  2. calyne
    • 2 years ago
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    yeah. this i know.

  3. calyne
    • 2 years ago
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    i got [2-ln(x-1)]/[(1-ln(x-1))^2] but that's wrong

  4. campbell_st
    • 2 years ago
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    I think it looks more like this u = x du/dx = 1 v = 1-ln(x -1) dv/dx = -1/(x-1) \[dy/dx = (1 - \ln(x -1) + x/(x-1))/(1-\ln(x -1))^2\]

  5. calyne
    • 2 years ago
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    The answer in the texbtook is f'(x) = [2x - 1 - (x - 1) ln(x - 1)] / [(x - 1)[1 - ln(x - 1)]^2]; domain of f = (1, 1+e) U (1 + e, infinity)

  6. calyne
    • 2 years ago
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    I need this explained to me.

  7. calyne
    • 2 years ago
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    still.

  8. bmp
    • 2 years ago
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    Optionally, use the reverse multiplication rule. Say: f(x) = x*(1-ln(x-1))^(-1) Quotient rule tend to be quite messy. But can you show what you did so far? I may be able to guide you through it. :-)

  9. myininaya
    • 2 years ago
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    \[y=\frac{x}{1-\ln(x-1)}\] You could do log diff. \[\ln(y)=\ln(\frac{x}{1-\ln(x-1)})\] \[\ln(y)=\ln(x)-\ln(1-\ln(x-1))\] Now diff both sides \[\frac{y'}{y}=\frac{1}{x}-\frac{0-\frac{1}{x-1}}{1-\ln(x-1)}\] So Simplifying right hand side first we have \[\frac{y'}{y}=\frac{1}{x}-\frac{-1}{(x-1)(1-\ln(x-1))}\] Now multiply y on both sides \[y'=y(\frac{1}{x}+\frac{1}{(x-1)(1-\ln(x-1))})\] And replace y with \[\frac{x}{1-\ln(x-1)}\] So this is another way :)

  10. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=differentiate+x%2F%5B1-ln%28x-1%29%5D good old classic way click on <show steps> to take a peek at the steps.

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