Connection between Fréchet derivative and the directional derivative in finite euclidean space In the lecture notes I am reading, the following statement is made: 
Let $U$ be an open subset of $R^n$, and define the function $e:U \to R$. $e$ is said to be differentiable if for every $u \in U$ there exists a linear functional $u' \in (R^n)'$ (The dual space of $R^n$) such that:
$$
\underset {||{h}|| \to 0}{\lim} \frac{e(u+h) - e(u) - \langle u', h \rangle}{ || h||} =0
$$
The derivative is then the function, $e' : U \to (R^n)'$ that maps a unique linear functional in the dual space to each vector in $U$ for which the limit is true.
Then the directional derivative in direction $h \in R^n$ is admitted by e if the following limit exists:
$$
\frac{\partial e}{ \partial h} (u) := \underset{t \to 0}{\lim} \frac{e(u + t h) - e(u)}{t}
$$
The notes go on to make this statement, which I am not sure how they got:
If $e$ is differentiable, then it admits at every $u \in U$ directional derivatives in all directions $h \in R^n$ and:
$$
\frac{\partial e}{ \partial h} (u) = e'(u) h
$$
Can anyone point me in the direction of a proof for this statement?
 A: It may be of some value to write an explicit answer (which is pretty similar to the comment by user251257, only containing more details).
Let $e:U\to\mathbb{R}$ be differentiable, and let $e':U\to(\mathbb{R}^n)^*$ denote the differential of $e$. Let $u\in U$, and let $h\in\mathbb{R}^n$. By differentiability, we have$$\lim_{t\to0}\frac{e(u+th)-e(u)-e'(u)(th)}{\|th\|}=0.$$By linearity of $e'(u)$, and by $\|th\|=|t|\|h\|$, we deduce$$\lim_{t\to0}\frac{e(u+th)-e(u)-te'(u)(h)}{t}=0,$$or alternativily,$$\lim_{t\to0}\frac{e(u+th)-e(u)}{t}=e'(u)(h).$$ 
A: The directional derivative is just the derivative of the composition:
\begin{alignat*}{2}
I\subset\mathbf R&\to U\subset\mathbf R^n &&\to\mathbf R\\
t&\mapsto u+th&&\mapsto e(u+th)
\end{alignat*}
where the interval $I$ is such that $u+th\in $ for all $t\in I$. Now observe that for a function of one variable, differentiability and derivability are equivalents. In particular, $t\mapsto u+th$ is differentiable, and its differential is but its linear part. In general, the differential of a one-variable function at $t=t_0$ is multiplication by the derivative at $t_0$.
Thus the directional derivative exists because the composition of these functions is a composition of differentiable functions. Furthermore, the chain rule for differentials says the differential of this one-variable function is multiplication by $\langle e',h\rangle$, or, what  amounts to the same, the derivative is $\langle e',h\rangle$.
