Question

Cum pot sa l determin pe a ?

Answer 1

Salut,

$F'(x)=(ln\sqrt{1+x^2})'}=\dfrac{(\sqrt{1+x^2})'}{\sqrt{1+x^2}}=\dfrac{\dfrac{1}2\cdot\dfrac{2x}{\sqrt{1+x^2}}}{\sqrt{1+x^2}}=\\\\=\dfrac{x}{1+x^2},\ deci\ a=0.\ Simplu,\ nu\ ?\ :-))).$

Green eyes.