Doubts about directional derivative in one variable Sometimes I read that directional derivatives for function in one variables are right and left derivatives. But this doesn't make sense to me. The only unit vector in $\mathbb{R}$ are $\pm1 $ , so we are saying that:
$$D_1f(x)=f'_+(x)=\lim_{t\to 0^+} \frac{f(x+t)-f(x)}{t}$$
$$D_{-1}f(x)=f'_-(x)=\lim_{t\to 0^-} \frac{f(x+t)-f(x)}{t}$$
But by definition:
$$D_{\mathbf{v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$
By this logic:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}-t\mathbf{v})-f(\mathbf{x})}{t}$$
I can substitute $-t=u$ ($-t$ doesn't assume infinite times the value $0$ so I can apply composite function limit theorem), so:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{-u}=-\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{u}=-D_{\mathbf{v}} f(\mathbf{x})$$
This should mean that:
$$f'_+(x)=-f'_-(x)$$
That in general is false!
Wouldn't be more correct to say that by convention in one variable we use derivative along the unit vector $1$ and that right and left derivatives are simply the right and left directional derivatives along the unit vector $1$.
Thanks in advance.
 A: You are right. Left & right derivatives and directional derivative are different notions – they coexist in fact, also in one-dimensional spaces such as $\mathbf{R}$.
The directional derivative can be defined on a generic vector- or affine space or on a differential manifold, without the existence of a norm or scalar product. It's therefore useful not only in many dimensions, but also in one dimension when the notion of "unit vector" is undefined.
More precisely we define, at a point $\pmb{x}$, the right directional derivative $\partial_{\pmb{v}^+}f$ of a function along the vector $\pmb{v}$ as
$$\lim_{t\to 0+} \frac{f(\pmb{x}+t\pmb{v}) - f(\pmb{x})}{t}$$
or equivalently
$$\lim_{n\to +\infty} \frac{f(\pmb{x}+\pmb{v}/n) - f(\pmb{x})}{1/n}.$$
The left directional derivative is defined identically except for $t \to 0-$ or $n\to -\infty$. From these definitions we have $\partial_{\pmb{v}^+}f \equiv -\partial_{-\pmb{v}^-}f$ (so I'm not sure about your last statement "this in general is false" – it depends on how you're defining $f_{\pm}'(x)$). Note how this limit doesn't involve any scalar product or norm in the space of $\pmb{v}$.
If both left and right directional derivatives along $\pmb{v}$ exist and are the same, then we just speak of the directional derivative along $\pmb{v}$. In differential geometry it is also denoted simply as "$\pmb{v}(f)$". We can have an interesting interplay of these two notions on the boundary of a manifold with boundary, even a 1D one, where it can happen that both $\pmb{v}$ and $-\pmb{v}$ can be defined at the boundary and yet we can only speak of the left or right derivative with respect to one or the other.
Intuitively and informally speaking, the directional derivative tells us something about the rate of change of a function as we move on the vector/affine space or manifold along some direction with some velocity, represented together by the vector $\pmb{v}$. The left and right derivatives tell us something about possible discontinuities in such rate of change.
So no, it isn't true that "directional derivatives for function in one variables are right and left derivatives", but we can say that the left derivative is minus the right derivative with respect to $-\pmb{e}$, where $\pmb{e}\equiv 1$ is the canonical unit vector, and vice versa.
For derivatives on vector spaces and differential manifolds, also with boundary, see eg Choquet-Bruhat, DeWitt-Morette, Dillard-Bleick's Analysis, Manifolds and Physics. Part I: Basics (rev. ed., Elsevier 1996), and Curtis, Miller's Differential Manifolds and Theoretical Physics (Academic Press 1985). There are of course many other good books out there on these topics.
A: The directional derivative is not the same as the limit approaching from the left or the right.  If you use the negative unit vector, you're asking what the slope is if you're traveling to the right vs. to the left.  You get the negative of the slope if you use the negative unit vector.  But that is correct. The divide by the
magnitude of v ($|v|$) is superfluous if you restrict v to be a unit vector.
$$\nabla_v f(x) = \nabla f(x) . \frac{v}{|v|}$$
