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anonymous
 one year ago
derive y = ln(log(x))
anonymous
 one year ago
derive y = ln(log(x))

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it \[1 \over xlog(x)\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0log(x) can also be ln(x)?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2only when it's base e.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2And yes you are correct.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.2So you let \(u=\log(x) ~,~ du = \dfrac{1}{x}\), then \[\frac{d}{dx} (y=\ln(u)):\]\[y'= \frac{1}{u}\cdot u' = \frac{1}{x\log(x)}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[y = ln(log_b(x))\] \[e^ y = log_b(x) = \frac{ln \ x}{ln \ b}\] \[e^y.y' = \frac{1}{ln \ b}.\frac{1}{x}\] \[y' = \frac{1}{ln \ b}.\frac{1}{x}.\frac{1}{e^y} = \frac{1}{ln \ b}.\frac{ 1}{x.log_b \ x} \] \[= \frac{ 1}{ x \ ln x} , \ b = e\]
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