Consider $D \subset \mathbb{C}^{n + 1}$ be a hypersurface such that $D \cong T \times D_{0}$, where $T$ be a disc in $\mathbb{C}$ and $D_{0}$ be a hypersurface in $\mathbb{C}^{n}$ [for this isomorphim, see Vector fileds on analytic spaces - Hugo Rossi].
How do you pove the next isomorphism
$$ \Omega_{D}^{q} \cong \Big(\Omega_{D_{0}}^{q} \oplus \Omega_{D_{0}}^{q-1}\wedge dt \Big) \otimes \mathcal{O}_{\mathbb{C}},$$
where $\Omega_{D}^{q}$ denote the q-forms sheaf in $D$ and $t$ denote de variable of $\mathbb{C}$ in above decomposition.
Suppose $\Omega^1_D$ is locally free. Since $D = D_0 \times T$, $\Omega^1_D = \Omega^1_{D_0}\oplus \Omega^1_T$. Then the exterior algebra of $\Omega^1_D$ becomes $$ \bigwedge \Omega^1_D = \bigwedge \left(\Omega^1_{D_0} \oplus \Omega^1_T\right) = \bigwedge \Omega^1_{D_0} \otimes \bigwedge \Omega^1_T. $$ Therefore, it remains to take the homogeneous component of degree q: $$ \Omega^q_D = \bigwedge^q \Omega^1_D = \bigoplus_{i+j=q}\bigwedge^i \Omega^1_{D_0} \otimes \bigwedge^j \Omega^1_T, $$ and notice that the dimension of T is 1.
If $\Omega^1_D$ is only coherent I believe you can use the fact above with a suitable resolution. But I'm not sure...