Let $X$ be a smooth scheme and $D\subset X$ be a divisor on $X$. I want to compute the sheaves ${\mathcal Ext}^i_{\mathcal{O}_X}(\mathcal{O}_D, \mathcal{O}_X)$. Actually it is quite easy. We have the following locally free resolution of $\mathcal{O}_D$: $$0\to\mathcal{O}_X(-D)\stackrel{\varphi}\to\mathcal{O}_X\to\mathcal{O}_D\to0.$$ Applying ${\mathcal Hom}_{\mathcal{O}_X}(\_,\mathcal{O}_X)$ to the complex $$0\to\mathcal{O}_X(-D)\stackrel{\varphi}\to\mathcal{O}_X\to0$$ we obtain the complex $$0\to\mathcal{O}_X\stackrel{\varphi^*}\to\mathcal{O}_X(D)\to0,$$ so ${\mathcal Hom}_{\mathcal{O}_X}(\mathcal{O}_D,\mathcal{O}_X)=\text{ker}\,\varphi^*$, ${\mathcal Ext}^1_{\mathcal{O}_X}(\mathcal{O}_D, \mathcal{O}_X)=\mathcal{O}_X/\text{im}\,\varphi^*$.
Now, to finish the computations, how can I prove the injectivity of $\varphi^*$?
As Mariano Suárez-Alvarez suggested I can check it locally. Let $\varphi:I\to A$ be an inclusion of an ideal to the ring, where $I$ is prime ideal of codimension one. I have a map $$\varphi^*:Hom_A(A,A)→Hom_A(I,A)$$ which acts by sending $f\in Hom_A(A,A)$ to $\varphi^*(f):=f\circ\varphi∈Hom_A(I,A)$. To prove that $\varphi^*$ is an injection we need to check that $f\circ\varphi=0$ implies $f=0$. The condition $f\circ\varphi=0$ gives that $f(I)=0$. The last condition implies that $I\cdot a=0$, where $a=f(1)$. Thus $I\subset\text{ann}(a)$. What should I do next?