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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)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0only when it's base e.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And yes you are correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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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