Question

derive y = ln(log(x))

Answer 1

is it \[1 \over xlog(x)\]?

Answer 2

log(x) can also be ln(x)?

Answer 3

only when it's base e.

Answer 4

And yes you are correct.

Answer 5

thank you :)

Answer 6

No problem

Answer 7

So you let \(u=\log(x) ~,~ du = \dfrac{1}{x}\), then \[\frac{d}{dx} (y=\ln(u)):\]\[y'= \frac{1}{u}\cdot u' = \frac{1}{x\log(x)}\]

Answer 8

\[y = ln(log_b(x))\]
\[e^ y = log_b(x) = \frac{ln \ x}{ln \ b}\]
\[e^y.y' = \frac{1}{ln \ b}.\frac{1}{x}\]
\[y' = \frac{1}{ln \ b}.\frac{1}{x}.\frac{1}{e^y} = \frac{1}{ln \ b}.\frac{ 1}{x.log_b \ x} \]
\[= \frac{ 1}{ x \ ln x} , \ b = e\]

Answer 9

well then!