Definition of total and directional derivatives I'm reading Analysis II by Terence Tao and having a problem with solving one of the exercises. We are given two definitions: 

Let E be a subset of R (the real numbers), $f:E \rightarrow R^m$ be a function, $x_{0}\in E$, and let $f'(x_{0}): R^n \rightarrow R^m$ be a linear transformation. We say that f is differentiable at $x_{0}$ with derivative $f'(x_{0})$ if

$\lim_{x \to x_{0}, x \in E- \{x_{0}\}} \frac{|f(x)-(f(x_{0})-f'(x_{0})(x-x_{0}))|}{|x-x_{0}|}=0$, where $|(x_{1}, ... x_{n})| = \sqrt{x_{1}^2+ ... + x_{n}^2}$.

Let $v \in R^n$, and let $x_{0}$ be an interior point of E.

If the limit $D_{v}f(x_{0})=\lim_{t \to 0; t&gt;0, x_{0}+tv \in E}\frac{f(x_{0}+tv)-f(x_{0})} {t}$ exists, we say that
f is differentiable in the direction v at $x_{0}$.

Now I have to prove that if f is differentiable at $x_{0}$, then it is also differentiable in the direction v at $x_{0}$ and $D_{v}f(x_{0})=f'(x_{0})v$. Isn't, though, $f'(x_{0})$ a vector? How is that last equation defined?
Let's understand what the derivative really is.

Given a point $x_0\in E \subset \mathbb{R}^n$ we define de derivative of $f$ in $x_0$ as a linear transformation, then, the function that for each $x \in E$ outputs the derivative of $x$ is a function from $E$ to $\mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ (the set of all linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$):

\begin{align}
f' : E \to &amp; \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \\
x \mapsto &amp; f'(x) : \mathbb{R}^n \to \mathbb{R}^m\\
\end{align}

Thus $f'(x_0)$ is a linear transformation and then $f'(x_0)v$ is a well-defined vector.

It is interesting to think about the following fact to fully understand this definition: we always listen that the derivative of a linear transformation $T:\mathbb{R}^n\to\mathbb{R}^m$ is $T$, but, in fact, the derivative of a linear transformation is a constant function $T': \mathbb{R}^n \to \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ such that $T'(x) = T$ for all $x\in \mathbb{R}^n$.
We have

$$f(x_0+tv) - f(x_0) = f'(x_0)(tv) + o(tv)=tf'(x_0)(v) + o(t).$$

Dividing by $t$ then gives

$$\frac{f(x_0+tv) - f(x_0)}{t} = f'(x_0)(v) + o(1).$$

This gives the desired result, namely $D_v(x_0) = f'(x_0)(v).$