This question asked yesterday got me thinking. While the derivatives of the tangent function span an infinite dimensional vector space over $\mathbb{C},$ the transcendence degree of the field generated by these derivatives is finite.
Here are two ways to see this. One is to observe that
$X^2 - DX + 1 = 0$
where $X$ denotes the tangent function. While a better way is to observe that via the sum/product rule, the field obtained by adjoining the set of derivatives of a function $f$ to $\mathbb{C}$ denoted $\mathbb{C}(\mathcal{D}f)$ is closed under differentiation and $\mathbb{C}(\mathrm{tan}) \subset \mathbb{C}(Exp) = \mathbb{C}(\mathcal{D}Exp)$
In fact, using the basic rules of differentiation, one can show that if the transcendence degrees over $\mathbb{C}$ of $\mathbb{C}(\mathcal{D}f)$ and $\mathbb{C}(\mathcal{D}g)$ are finite so too are the transcendence degrees of $\mathbb{C}(\mathcal{D}(f + g)),$ $\mathbb{C}(\mathcal{D}(fg))$ and $\mathbb{C}(\mathcal{D}(f \circ g)).$
Which brings me to my question:
Do there exist any functions $f$ which are meromorhic on $\mathbb{C}$ such that the transcendence degree of $\mathbb{C}(\mathcal{D}f)$ over $\mathbb{C}$ is infinite?