Question

URGENT!!derivata va rog!!!

Answer 1

Răspuns:

(lnx/x)' =

[(lnx)'×(x) - (lnx)×(x)'] / (x^2) =

[(1/x)×(x) - (lnx)×(1)] / (x^2) =

(1-lnx)/(x^2)

Answer 2

Răspuns:

Explicație pas cu pas:

$\Big( \frac{lnx}{x} \Big)' = \frac{\Big(lnx\Big)' \cdot x - lnx \cdot \Big(x \Big)' }{x {}^{2} } = \frac{ \frac{1}{x} \cdot x - lnx \cdot 1 }{x {}^{2} } = \boxed{ \frac{1 - lnx}{x {}^{2} }}$