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Deriving ln[x^n]

Name: Anonymous 2005-12-01 15:02

So the mathbook says:
f(x) = ln[x^n]
f'(x) = n/x
but it doesn't give an explanation to why. Please do tell me as the website of my distance education company isn't responding at the moment.

Name: Anonymous 2005-12-01 15:40

sigh...

f  = log(x^n) = n*log(x)
f' = n*(log(x))'
f' = n*(1/x)
f' = n/x

Name: Anonymous 2005-12-01 16:55

Nevermind, I figured it out. I had forgotten that:
ln[x^n] = n * ln[x]

Name: Anonymous 2005-12-01 16:57

(Had the thread open in tab, wrote reply, added it, saw that you had already answered)

Thanks anyways. The logarithm stuff was in an earlier course so I had forgotten most of it.

Name: Anonymous 2005-12-02 13:44

Why the curly brackets? Curly brackets access arrays. Use parenthesis for functions and operations.

Name: Anonymous 2005-12-02 17:39

>>4

you'll never be good at maths if you forget essential shit like this. then again your goal might not be to be good at maths, in which case you suck

Name: Anonymous 2005-12-02 19:05

>>5
you mean square brackets. and who cares...they are all brackets anyway. this is not a programming language.

Name: Anonymous 2005-12-02 19:40

() = parentheses
[] = brackets
{} = braces

And very few people care.

Name: Anonymous 2005-12-03 15:18

>>6
Thanks mate.

Don't change these.
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