In Lemma 6.2.4 of his book, Qing Liu states
Let $ X \to S $ be a morphism of finite type over a locally Noetherian scheme $ S $. Fix $ s \in S $, $ x \in X_{s} $ and let $$ d = \dim_{k(x)} \Omega_{X_{s} /k(s) , x } \otimes _ { \mathcal{O}_{X_s} } k(x) . $$
My question is, isn't the module $ \Omega_{X_{s}/k(s), x } \otimes_{\mathcal{O}_{X_s} } k(x) $ the fiber of $ \Omega_{X/S} $ at $ x \in X $?
The reason I think that is that first I pullback the sheaf $ \Omega_{X/S} $ along the morphism $ X_{s} \to X $ to get $ \Omega_{X_{s} / k(s) } $ and then I pullback along the morphism $ k(x) \to X_{s} $ to get the sheaf/module on $ k(x) $. But since the $ k(x) $ is independent of the space $ X $ or the fiber $ X_{s} $, I have a composition $ k(x) \to X_{s} \to X $, and thus, I am pulling back along this composition.