Question

\[\frac{ d }{ dx } \left[ \ln (\ln x) \right]\]

Answer 1

do you know chain rule ?

Answer 2

yeah i know chain rule

Answer 3

so, did you try this ? what u got ?

Answer 4

so do i just find
\[\ln (1/x)\]

Answer 5

first we find derivative of outer function, then inner.
ok, let me give you an example :
if y = sin (ln x)
then y' = cos (ln x) * [d/dx (ln x)] = cos (ln x) *(1/x)
here, sin was outer function, so we have cos (ln x)
can u try now for ln (ln x ) ?

Answer 6

but isn't the outer function ln(x) as well?

Answer 7

here, yes.

Answer 8

what i gave was just another example.

Answer 9

You are misunderstanding the chain rule.

D(f(g(x)) is equal to f'(g(x))g'(x)

Answer 10

Not f'(g'(x))

Answer 11

i understand but then isn't it \[\frac{ 1 }{ x \ln x }*\frac{ 1 }{ x }\]

Answer 12

why the extra 1/x ?
1/x will come once only, right ? as d/dx (ln x)

Answer 13

isn't it f '(gx)*g'(x)

Answer 14

\((\ln (\ln x))' = (1/(\ln x))(\ln x)'= 1/ [x \ln x]\)

Answer 15

Let's try replacing the inner ln(x) with u

Answer 16

here, f' (g(x)) = 1/ ln x
because f'(x) would have been 1/x

Answer 17

es, if you are getting confused, try substitution method.

Answer 18

f'(ln(u))= u'/u where u=lnx

so u' = 1/x and u = ln x so

f' = 1/xlnx

Answer 19

oh I see I was basically doing a step twice

Answer 20

Correct

Answer 21

k..with some practice, you'll be able to do it mentally :)

Answer 22

thanks guys for taking the time out to help me. Especialy you hartnn because I see you have a lot of followers.