How do you find the derivatives of $s = ln \sqrt[t]{t}$ $s=lnroott(t)$ ?

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Answer 1 · 2017-03-19T15:16:27

$\frac{d}{d t} (ln \sqrt[t]{t}) = \frac{1 - ln t}{t^{2}}$ $d/(dt) ( ln root (t)(t)) =(1- lnt)/t^2$

Using the properties of logarithms:

$s = ln \sqrt[t]{t} = \frac{1}{t} ln t$ $s = ln root(t)(t) = 1/tlnt$

we can differentiate using the product rule:

$\frac{d s}{d t} = \frac{1}{t} \frac{d}{d t} ln t + ln t \frac{d}{d t} (\frac{1}{t})$ $(ds)/(dt) = 1/t d/(dt) ln t + lnt d/(dt)(1/t)$

$\frac{d s}{d t} = \frac{1}{t} \frac{1}{t} + ln t (- \frac{1}{t^{2}})$ $(ds)/(dt) = 1/t 1/t + lnt (-1/t^2)$

$\frac{d s}{d t} = \frac{1}{t^{2}} (1 - ln t)$ $(ds)/(dt) = 1/t^2(1- lnt)$

How do you find the derivatives of s=lnroott(t)s=lnt√ts=lnroott(t)?