Let $X$ be a smooth projective variety and let $D$ be an effective divisor on $X$. Is there a natural way to describe the tangent space to $D$ (or $D^\vee$, of course) at a divisor $D'$? Preferably as some sort of cohomology group, again ideally on $X$. I would prefer to avoid using the fact that the linear system is a projective space, and do things naturally.
The linear system $D$ is the projectivization of $H^0(X,\mathcal O(D)).$ The divisor $D'$ corresponds to a line $\ell_{D'} \subset H^0(X,\mathcal O(D)).$ (This line consists of all the sections whose zero loci are equal to $D'$.) The space $Hom(\ell_{D'},H^0(X,\mathcal O(D))/\ell_{D'})$ is a vector space of the same dimension as $D$, and is naturally isomorphic to the tangent space of $D$ at $D'$. (Here is am using the general fact that if $V$ is an vector space, and $\ell \subset V$ a line through the origin, then the tangent space to the projectivization $\mathbb P(V)$ at the point corresponding to the line $\ell$ is identified with $Hom(\ell,V/\ell)$.) [Thanks to Georges Elencwajg for correcting an earlier misstatement here.] One can say a little more; before doing so, it's convenient to note that $D$ can be any divisor in the linear system, and so it is no loss of generality to set $D = D'$; this eases the notation somewhat. We also fix a section of $\mathcal O(D)$ cutting out $D$, i.e. a basis of $\ell_D$, which gives an identification $\ell_D = k = H^0(X,\mathcal O).$; this allows us to rewrite $Hom(\ell_D,H^0(X,\mathcal O(D))/\ell_D)$ simply as $H^0(X,\mathcal O(D))/\ell_D$. Our choice of basis for $\ell_D$ gives a short exact sequence $$0 \to \mathcal O \to \mathcal O(D) \to \mathcal O(D)\_{ D} \to 0,$$ and taking global sections gives $$0 \to \ell_{D} \to H^0(X,\mathcal O(D)) \to H^0(D, \mathcal O(D)\_{ D}),$$ and hence an injection $$H^0(X,\mathcal O(D))/\ell_{D} \hookrightarrow H^0(D,\mathcal O(D)\_{ D}).$$ But this is not going to be an isomorphism in general, I guess. Indeed, the cokernel embeds into $H^1(X,\mathcal O)$, which is the tangent space to Pic $X$, while $H^0(D,\mathcal O(D)\_{D})$ is the tangent space to the Hilbert scheme of $X$ at $D$. [Note: To see this, observe that our choice of section cutting out $D$ corresponds to a choice of isomorphism $\mathcal O(D) \cong \mathcal I_D^{1},$ and it is $(\mathcal I_{D}^{1})\_{ D}$ that is canonically the normal bundle to $D$.] The map $H^0(D,\mathcal O(D)_{D}) \to H^1(X,\mathcal O)$ then measures the extent to which the deformations of $D$ in $X$ fill up the component of the Picard scheme containing $D$. I imagine that if $D$ is sufficiently positive then this map is surjective; at least when $X$ is a surface, this is the main result of Mumford's "Lectures on curves on an algebraic surface" (if I am remembering correctly). 

