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calyneBest ResponseYou've already chosen the best response.0
i got [2ln(x1)]/[(1ln(x1))^2] but that's wrong
 2 years ago

campbell_stBest ResponseYou've already chosen the best response.1
I think it looks more like this u = x du/dx = 1 v = 1ln(x 1) dv/dx = 1/(x1) \[dy/dx = (1  \ln(x 1) + x/(x1))/(1\ln(x 1))^2\]
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
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)
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
I need this explained to me.
 2 years ago

bmpBest ResponseYou've already chosen the best response.0
Optionally, use the reverse multiplication rule. Say: f(x) = x*(1ln(x1))^(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. :)
 2 years ago

myininayaBest ResponseYou've already chosen the best response.1
\[y=\frac{x}{1\ln(x1)}\] You could do log diff. \[\ln(y)=\ln(\frac{x}{1\ln(x1)})\] \[\ln(y)=\ln(x)\ln(1\ln(x1))\] Now diff both sides \[\frac{y'}{y}=\frac{1}{x}\frac{0\frac{1}{x1}}{1\ln(x1)}\] So Simplifying right hand side first we have \[\frac{y'}{y}=\frac{1}{x}\frac{1}{(x1)(1\ln(x1))}\] Now multiply y on both sides \[y'=y(\frac{1}{x}+\frac{1}{(x1)(1\ln(x1))})\] And replace y with \[\frac{x}{1\ln(x1)}\] So this is another way :)
 2 years ago

experimentXBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=differentiate+x%2F%5B1ln%28x1%29%5D good old classic way click on <show steps> to take a peek at the steps.
 2 years ago
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