Tangent space of a point of an algebraic variety Let $V$ be a non-singular affine variety in $\mathbb{C}^n$.
$V$ can be regarded as a complex manifold.
Let $p = (a_1,\dots,a_n) $ be a point of $V$.
Let $\mathcal{O}_p$ be the local ring of $V$ at $p$.
A tangent vector $v$ at $p$ is a derivation $\mathcal{O}_p \rightarrow \mathbb{C}$, i.e.
a $\mathbb{C}$-linear map $v$ such that $v(fg) = v(f)g(p) + f(p)v(g)$ for $f, g \in \mathcal{O}_p$.
Let $T_p$ be the set of tangent vectors at $p$.
We regard $T_p$ as a vector space over $\mathbb{C}$ in the obvious way.
On the other hand, we can define a tangent space at $p$ as follows.
Let $f_1,\dots f_r$ be defining polynomials for $V$.
Let $L_i$ be the hyperplane defined by $\sum_k \frac{\partial f_i}{\partial x_k}(p)(x_k - a_k) = 0$.
Let $S_p = \bigcap_i L_i$.
Are $T_p$ and $S_p$(or rather the vector space attached to it) canonically isomorphic?
 A: Lemma 1
Let $A$ be a local algebra over a field $k$.
Let $\mathfrak{m}$ be the unique maximal ideal of $A$.
Suppose the canonical homomorphism $k \rightarrow A/\mathfrak{m}$ is an isomorphism.
Let $f \in A$.
There exists a unique $c \in k$ such that $f \equiv c$ (mod $\mathfrak{m}$).
We denote this $c$ by $f(0)$.
We call a $k$-linear map $v\colon A \rightarrow k$ a derivation if $v(fg) = v(f)g(0) + f(0)v(g)$ for $f, g \in A$.
Let $Der(A, k)$ be the set of derivations.
We regard $Der(A, k)$ as a vector space over $k$ in the obvious way.
Let $v \in Der(A, k)$.
If $f, g \in \mathfrak{m}$, $v(fg) = v(f)g(0) + f(0)v(g) = 0$.
Hence $v(\mathfrak{m}^2) = 0$.
Hence $v$ induces a $k$-linear map $\bar v\colon \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$.
Hence we get a $k$-linear map $\psi\colon Der(A, k) \rightarrow (\mathfrak{m}/\mathfrak{m}^2)^*$.
Then $\psi$ an isomorphism.
Proof:
This is proved here
Lemma 2
Let $k$ be a field.
Let $L_i = \sum_{k = 1}^{n} a_{ik} x_k, i = 1,\dots, r$ be linear polynomials in $k[x_1,\dots,x_n]$.
Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$.
Let $W$ be the vector subspace of $k^n$ generated by $L_i(e_1,\dots,e_n)$ for all $i$, where $\{e_1,\dots,e_n\}$ is the canonical basis of $k^n$.
Then $T$ is canonically isomorphic to $(k^n/W)^*$.
Proof:
An element of $(k^n/W)^*$ is identified with a linear map $f\colon k^n \rightarrow k$ such that $f(W) = 0$. Such a map $f$ is uniquely determined by the condition $\sum_{k = 1}^{n} a_{ik} f(e_k) = 0, i = 1,\dots, r$.
Hence the assertion follows.
QED
Proposition
Let $A = k[x_1,\dots,x_n]$ be the polynomial ring over $k$.
Let $I$ be an ideal of $A$ generated by $F_1,\dots,F_r$.
Let $B = A/I$.
Let $p = (a_1,\dots, a_n)$ be a point of $k^n$ such that $F_i(p) = 0$ for all $i$.
Let $M = (x_1 - a_1,\dots,x_n - a_n)$ be the maximal ideal of $A$.
Let $\mathfrak{m} = M/I$.
Let $L_i = \sum_k \frac{\partial F_i}{\partial x_k}(p)x_k$ for $i = 1, \dots, r$.
Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$.
Then $T$ is canonically isomorphic to $Der(B_{\mathfrak{m}},k)$.
Proof:
$\mathfrak{m}/\mathfrak{m}^2$ is canonically isomorphic to $M/(I + M^2)$.
$M/M^2$ is a $k$-vector space with a basis $\bar x_1,\dots,\bar x_n$, where $\bar x_i = x_i$ (mod $M^2$).
Let $F \in I$.
Since $F(p) = 0$, $F = \sum_k \frac{\partial F}{\partial x_k}(p)(x_k - a_k) + \cdots$.
Hence $(I + M^2)/M^2$ is a vector subspace of $M/M^2$ generated by $L_i(\bar x_1, \dots,\bar x_n)$ for all $i$.
Hence the assertion follows from Lemma 1 and Lemma 2.
QED
A: Yes, there is a canonical isomorphism of $\mathbb C$-vector spaces $$i:S_{X,p}\stackrel {\cong}{\to} T_{X,p}=Der(\mathcal O_{X,p}, \mathbb C)$$ If one starts with a vector $v=(v_1,...,v_n)\in S_p\subset \mathbb C^n$ (so that $\sum_k \frac{\partial f_i}{\partial x_k}(p)v_k = 0$), the isomorphism $i$  associates to it the derivation $i(v)=\partial _v$ defined on $\mathcal O_{X,p}$ by the formula $$        \partial_v(g)=\sum_k \frac{\partial g}{\partial x_k}(p)v_k $$ where $g\in \mathcal O_{X,p}$ is an arbitrary local function.
The inverse isomorphism is given by $$i^{-1}: Der(\mathcal O_{X,p},\mathbb C) \stackrel {\cong}{\to} S_{X,p} :\partial \mapsto (\partial x_1,...,\partial x_n)             $$
Edit
 The vector space of derivations is also isomorphic to Zariski's tangent space $(\frak m_p/\frak m^2_p)^*$, defined via the maximal  ideal $\mathfrak m_p\subset \mathcal O_{X,p}$.
 The isomorphism is  $$ Der(\mathcal O_{X,p} )   \stackrel {\cong}{\to}  (\frak m_p/\frak m^2_p)^*:\partial \mapsto \overline{\partial}    $$ where $\overline{\partial} (g \;\text {mod}  \;m^2_p)=\partial (g)$ for $g\in \frak m_p$.  
Second Edit
The following remark may be of some interest, since it does not seem to be addressed in Algebraic Geometry books:
If $X\subset \mathbb A^n_k$ is an affine algebraic variey and if $p\in X$, we may consider the maximal ideal $M_p\subset \mathcal O(X)$ of global functions vanishing at $p$.
We may also consider, as we already did,  the maximal ideal $\mathfrak m_p\subset \mathcal O_{X,p}$ of germs of functions regular ay $p$ and vanishing at $p$.
We then have a natural $k$-linear map $ M_p/M_p^2       \to \mathfrak m_p/\mathfrak m_p^2         $ and the slightly surprising but pleasant fact is that this linear map is an isomorphism.
