I Is dfx(y) a dual vector due to linear transformation properties?

[LagrangeEuler]

f(\vec{x}+\epsilon \vec{y})-f(\vec{x})=\epsilon \mbox{d}f_{\vec{x}}(\vec{y})+O(\epsilon^2).
Is $\mbox{d}f_{\vec{x}}(\vec{y})$ dual vector and why? Is it because $\mbox{d}$ is linear transformation? Also why equality
f(\vec{x}+\epsilon \vec{y})-f(\vec{x})=\epsilon \mbox{d}f_{\vec{x}}(\vec{y})+O(\epsilon^2)
is correct?

 

[martinbn]

To get a meaningful answer you need to provide more context.

The equality is correct because that is how $df$ is defined. And it is a dual vector because it takes vectors as arguments and gives a number as a result, and it is linear in the argument (the $\vec{y}$ in you expression).