If $Y = X \backslash Z(f)$ for some affine variety $X$ with $p \in Y$, then $T_p Y \cong T_p X$ Let $Y = X \backslash Z(f)$ be a quasi-affine variety for some affine variety $X$, and let $p \in Y$. I'd like to prove that $T_p Y \cong T_p X$.
I have the following definition of $T_p X$:

If $A$ is a $k$-algebra and $\phi : A \to k$ a $k$-algebra homomorphism, then a $k$-linear map $D : A \to k$ is a 'derivation centred at $\phi$' if $D(fg) = \phi(f)D(g) + D(f)\phi(g)$ for all $f,g \in A$. Let $\mathrm{Der}(A,\phi)$ be the vector space of derivations centred at $\phi$. Define $T_p X = \mathrm{Der}(k[X], \mathrm{ev}_p)$.

My thoughts so far:
A morphism $\psi : Y \to X$ of affine varieties gives rise to a $k$-algebra homomorphism $\psi^*: k[X] \to k[Y]$ which maps $g \mapsto g \circ \psi$. This in turn gives rise to a $k$-linear map $d_\psi : T_pY \to T_{\psi(p)}X$, by sending $D \mapsto D \circ \psi^*$. 
So for the case in question, we have a morphism $i : Y \to X$ given by the inclusion map, under which $i(p) = p$, so this looks pretty promising. I need to show that the map $d_i$ is  a bijection. 
If $d_i(D) = 0$, then $D \circ i^* = 0$ - I think this can only happen if $D$ is the zero map, so we have injectivity. Surjectivity I'm unsure about, and I haven't used $f$ anywhere yet. Any hints would be appreciated
Thanks
 A: The variety $X$ has a local ring $\mathcal O_{X,p}$ at $p$.
If $\mathfrak m \subset\mathcal O_{X,p} 
 $ is its  maximal ideal we have $T^*_p(X)=\mathfrak m/\mathfrak m^2$, a $k$-vector space, and the dual of that vector space is the tangent space of $X$ at $p$ : $T_p(X)=(\mathfrak m/\mathfrak m^2)^*$.
Since the local ring $\mathcal O_{X,p}$ does not change if you replace $X$ by an open neighbourhood $U$ of $p$, you have $T_p(X)=T_p(U)$.
Apply to $U=X\setminus Z(f)=Y$  
The tangent space I defined is called the Zariski tangent space and is easily seen to be isomorphic  to your  tangent space defined with derivations.
The basic idea is that given your derivation $D$, you can extend it to $\mathcal O_{X,p}$, then restrict it to $\mathfrak m$ and since this linear form is zero  on $\frak m^2$ you finally obtain  a linear form on  $\mathfrak m/\mathfrak m^2$, a tangent vector in the Zariski sense.
A: In your definition of the tangent space, $X = \text{Spec }A$? I would approach this a little differently. Namely, show that any such derivation arises from a derivation of the local ring at $p$. It will then follow that your definition of the tangent space is independent of the choice of affine neighborhood of $p$ that you choose.
Note that a priori your definition of the tangent space doesn't make sense for non-affine varieties, so you really need to check independence of the choice of affine neighborhood anyway to have a reasonable definition. 
