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#1 2005-10-16 06:13:41
- Jaswahhihi
- Member
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Differentiation
How do you differentiate:
1 / ln x ????????????????????????????????????????????
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#2 2005-10-16 06:40:02
- mathsyperson
- Moderator
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Re: Differentiation
The quotient rule says that (u/v)' = (vu' - uv')/v², 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)².
There may be a simpler way of doing that that gives a simpler answer, but I'm sleepy so I can't see one.
Why did the vector cross the road?
It wanted to be normal.
#3 2005-11-08 21:41:07
- Ash
- Guest
Re: Differentiation
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...)