Question

Se considera  functia  f:(0,+ infinit)⇒R  f(x)=lnx-1/x

a) Aratati ca  f derivat (x)=$\frac{x+1}{ x^{2} }$  x∈(0,+ infinit)

Answer 1

[tex]f'(x)=(lnx- \frac{1}{x} )'=(lnx)'-( \frac{1}{x} )'= \frac{1}{x} - \frac{1'\cdot x-1\cdot x'}{x^2}\\ f'(x)=^{x)} \frac{1}{x} + \frac{1}{x^2}= \frac{x}{x^2}+\frac{1}{x^2}= \frac{x+1}{x^2} [/tex]