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(lnx - 1/x) totul derivat, iar x0 = 1

Răspuns :

[tex] \it \left(lnx-\dfrac{1}{x}\right)' =(lnx)'-\left(\dfrac{1}{x}\right)'= \dfrac{1}{x} -\left(\dfrac{-1}{x^2}\right)=\dfrac{1}{x} +\dfrac{1}{x^2}=\dfrac{x+1}{x^2} [/tex]