The exterior derivative is a map $\text d:\Omega^k\rightarrow\Omega^{k+1}$. We can divide two differential forms to get the ordinary derivative:
$$\dfrac {\text df}{\text dx}=D_xf.$$
Is there an analogous map for partial derivatives? A map $\partial$ such that
$$\dfrac {\partial f}{\partial x}=\partial_xf?$$
Only when you're on a one-dimensional manifold does it make sense to divide two $1$-forms. When you're on $\Bbb R$, in particular, since $T^*_p\Bbb R \cong \Bbb R$, we have $df_p = \lambda dx_p$ for some $\lambda$, and we can write $\dfrac{df}{dx}(p) = \lambda$. You can also look at it this way. Pick any tangent vector $v\in T_p\Bbb R$. Then $\lambda = df_p(v)/dx_p(v)$. Notice that this is independent of $v$, since $df_p(tv)/dx_p(tv) = df_p(v)/dx_p(v)$.
But when you're on a higher-dimensional manifold, it makes no sense to divide $1$-forms. Let's just try $\Bbb R^2$. What would it mean to take $$\frac{\omega}{\eta} = \frac{\alpha\, dx+\beta\, dy}{\gamma\, dx +\delta\, dy}?$$ Of course if it happens that $\omega = \lambda\eta$ for some scalar $\lambda$, then, sure, it works like before. What otherwise, what does it mean? We could try takine $\omega(v)/\eta(v)$, but this will change when we change $v$. (E.g., in the example above, if we take $v=e_1$, we get $\alpha/\gamma$, and if we take $v=e_2$, we get $\beta/\delta$.
With regard to your question, $d$ is coordinate-independent, whereas partial derivatives are obviously very dependent on your coordinate system. If you're on $\Bbb R^2$, you might want to rephrase your question in this fashion: Is there an operator $d_1\colon\Omega^0(\Bbb R^2)\to\Omega^1(\Bbb R^2)$ so that $d_1f = \dfrac{\partial f}{\partial x}dx$? This is completely coordinate-system dependent, so there's no invariantly-defined notion. (There is such a creature, for example, if we work on $\Bbb C$ and have the operators $\partial = \dfrac{\partial}{\partial z}dz$ and $\bar\partial = \dfrac{\partial}{\partial\bar z}d\bar z$, and these are invariantly defined if we make a holomorphic change of coordinates.)