Question

f(x) = ln x-(1/x). Aratati ca f'(x) = x+1/x^2, x E ( 0, +inf) ​

Answer 1

$\it f(x)=lnx-\dfrac{1}{x}\\ \\ \\ f'(x)=(lnx)'-(\dfrac{1}{x})'=\dfrac{1}{x}-(-\dfrac{1}{x^2})=\dfrac{^{x)}1}{\ x}+\dfrac{1}{x^2}=\dfrac{x+1}{x^2}$