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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2005-11-08 21:41:07

simpler way is probably using the chain rule. Take up logex to be (logex)^-1, then simply differentiate as normal:

-1(logex)^-2 x (1/x)

= -1/(x(logex)^2)

Sorry if my terminology isn't clear- im in australia and the syllabus is different here (just had my finals and im depressed over a poor effort in my most important exam...trying to find some solace by going over everything and brushing up my math skills...just can't believe how much effort i put in to make so many silly mistakes on the day...)

2005-10-16 06:40:02

The quotient rule says that (u/v)' = (vu' - uv')/v, where u' and v' are the differentials of u and v with respect to x.

Taking u = 1 and v = ln x means that u' = 0 and v' = 1/x, so (u/v)' would be -1/x(ln x).
There may be a simpler way of doing that that gives a simpler answer, but I'm sleepy so I can't see one.

2005-10-16 06:13:41

How do you differentiate:

1 / ln x ????????????????????????????????????????????

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