Why isn't the directional derivative with respect to $-\vec v$ the same as the directional derivative with respect to $\vec v$? Suppose $f$ is a differentiable function from $\mathbb R^2$ to $\mathbb R$, $\vec v$ is a unit vector in $\mathbb R^2$, and $\vec a$ is a point in $\mathbb R^2$.
Then, visually, taking the directional derivative of $f$ with respect to $\vec v$ at $\vec a$ represents taking the vertical plane containing $\vec v$, translating it to $\vec a$, intersecting it with the graph of $f$ to obtain a curve, and then finding the slope of the tangent line to that curve at $\vec a$.
But vertical plane containing $\vec v$ is the same as the vertical plane containing $-\vec v$, so why isn't the directional derivative of $f$ with respect to $-\vec v$ at $\vec a$ equal to the directional derivative of $f$ with respect to $\vec v$ at $\vec a$?
 A: The directional derivative at $a$ in direction $v$ is the rate at which the function is changing as you move from $a$ in that direction at unit speed. If the function is increasing as you move that way it will be decreasing when you move the other way.
You can see that change of sign if you write the directional derivative as the limit of the difference quotient.
A: Your description of the visual interpretation is not quite complete. It's not enough to just cut the graph with a vertical plane and start talking about the slope of the resulting curve. You also need to specify from which side you are supposed to look at the plane, since what from one side looks like a positive slope will look like a negative slope when viewed from the other side. Can you now see how this difference is connected to the difference in $\vec{v}$ vs. $-\vec{v}$?
A: When it comes to derivatives, then it is important to clarify what you mean by them, i.e. what the variable is: location, function, Jacobi matrix, direction, slope? Let's consider the Weierstraß formula:
$$
f(x_{0}+v)=f(x_{0})+J_{x_{0}}(v)+r(v)
$$
We see here, that the Jacobi matrix, is invariant of the direction. However, the slope $J_{x_{0}}(v) \to J_{x_{0}}(-v)=-J_{x_{0}}(v)$ changes the sign.
