The relationship between the differential and the directional derivative of a function I am currently studying differential manifolds (from John M. Lee 's book), and have a question concerning the difference between what is defined as the $\textbf{differential of a function F}$, and the $\textbf{directional derivative of a function F}$. Let $M \subset \mathbb{R}^{m}$, let $N\subset \mathbb{R}$, and suppose $F:M \rightarrow N$ is a smooth map. Then,


*

*The differential of $F$ at $p \in M$ is a map
$dF_{p}:T_{p}M \rightarrow T_{F(p)}N$
defined as, for some $v \in T_{p}M$, $dF_{p}(v)$ is a derivation in $T_{F(p)}N$ defined as, for all $f \in C^{\infty}(N)$, 
$dF_{p}(v)(f)=v(f \circ F)$. 
Now, because $M$ and $N$ are Euclidean themselves, if $v=(v_{1},...,v_{N})$, this can be expressed as 
$dF_{p}(v)(f)=v_{1} \cdot \frac{\partial f}{\partial F}\ \frac{\partial F}{\partial x_{1}}+...+v_{m} \cdot \frac{\partial f}{\partial F}\ \frac{\partial F}{\partial x_{m}}$. 


*The directional derivative of $F$ at $p$ in direction $v \in \mathbb{R}^{m}$ is given by 


$D_{v}F(p)=v_{1} \cdot \frac{\partial F(p)}{\partial x_{1}}+...+v_{m}\cdot \frac{\partial F(p)}{\partial x_{m}}$.
Now, it seems that if $Id:\mathbb{R}\rightarrow \mathbb{R}$ is the identity function on $\mathbb{R}$, we have that for $v \in T_{p}M$,
$v(F)=dF_{p}(v)(Id)=D_{v}F(p)$. 
Am I reading this correctly? I am trying to weed through the abstraction of differentials between manifolds and ground it into something more familiar, the directional derivative. Are directional derivatives in the theory of manifolds expressed as the the differential evaluated at the identity function, which are equivalent to simply evaluating $v(F)$ itself? 

The relationship between the differential and directional derivative is the same in differential manifolds as in Euclidean space. The derivative is a linear function. Linear functions take in vectors and output vectors. When the input vector is a unit vector, the output is called the directional derivative. That is because directions are defined by unit vectors.
Let's use an example.
Let $f:\mathbb R^2\to\mathbb R^2$ be defined by $f(x,y)=(x^2-y^2,xy)$
So $f=(f_1,f_2)$ where $f_1(x,y)=x^2-y^2$ and $f_2(x)=xy$
The derivative, or differential, is $df= \begin{pmatrix}
    \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
    \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\    \end{pmatrix}
= \begin{pmatrix}
    2x & -2y \\
    y & x \\    \end{pmatrix}$ 
If we want the derivative of $f$ at say $(1,2)$ then we get $$df(1,2) = \begin{pmatrix}
    2 & -4 \\
    2 & 1 \\    \end{pmatrix}$$
Imagine being at point $(1,2)$ in the plane. There are $360^{\circ}$ of directions. We represent each direction as a unit vector in that direction. Now we can ask questions like, what is the derivative of $f$ in the $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ direction at $(1,2)$?.
We evaluate the derivate to get $$\begin{pmatrix}
    2 & -4 \\
    2 & 1 \\    \end{pmatrix} \begin{pmatrix}
    \frac{1}{\sqrt{2}} \\
    \frac{1}{\sqrt{2}} \\    \end{pmatrix}=\begin{pmatrix}
    \frac{-2}{\sqrt{2}} \\
    \frac{-3}{\sqrt{2}} \\    \end{pmatrix}$$
