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Topic review (newest first)
- Ash
- 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...)
- mathsyperson
- 2005-10-16 06:40:02
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.
- Jaswahhihi
- 2005-10-16 06:13:41
How do you differentiate:
1 / ln x ????????????????????????????????????????????