Question

Is my result of d/dx ln(ln(ln(x))) correct ?


1/((ln(ln(x)))^2*ln(ln(ln(x)))*ln(x)*x)

Answer 1

very wierd problem ...

Answer 2

well,

d/dx ln( f(x) ) = 1/f(x) • f'(x) = f'(x)/f(x)
So, that is the princple, so let's see...

Answer 3

d/dx ln(ln(ln(x))) = 1/ln(ln(x)) • 1/(ln(x)) • 1/x

So,
d/dx ln(ln(ln(x))) = 1/\(\{\)x• ln(x) • ln[ln(x)]\(\}\)

is my result

Answer 4

is it right?
http://www.wolframalpha.com/input/?i=d%2Fdx+ln%28ln%28ln%28x%29%29%29
probably not

Answer 5

yup, that is right inkyvoyd

Answer 6

I can do step by step if you want but you probably want to look over your work.. looks like you differentiated one iteration too many or the likes.

Answer 7

and if you had

d/dx \((\)ln(ln(ln x))\()^n\)

then, you derivative would be
d/dx (ln(ln(ln(x)))\(^n\) =\(\{\) n•ln(ln(ln(x)))\(^{n-1}\)\(\}\)\(/\)\(\{\){x• ln(x) • ln[ln(x)]\(\}\)

Answer 8

I don't know how you got a ^2 in there tho

Answer 9

you just do a chain rule like that, so then...