Let $E\_2$, $E\_4$, and $E\_6$ denote the standard Eisenstein series.
The usual variables $q=e^{2\pi i\tau}$ allow us to regard the
$E\_n$'s as functions on either the upper half plane or the unit
disk and we can define $E\_n'=\frac{1}{2\pi
i}\frac{d}{d\tau}E\_n(\tau)=q\frac{d}{dq}E\_n(q)$. I had cause to
calculate a few of these and saw
$$ E\_4'=\frac{1}{3}(-E\_6+E\_4E\_2) $$
$$ E\_4''=\frac{5}{36}(E\_8-2E\_6E\_2+E\_4E\_2^2)$$
$$E\_4^{(3)}=\frac{5}{72}(-E\_{10}+3E\_8E\_2-3E\_6E\_2^2+E\_4E\_2^3) $$
$$E\_4^{(4)}=\frac{35}{864}(E\_4^3-4E\_{10}E\_2+6E\_8E\_2^2-4E\_6E\_2^3+E\_4E\_2^4)-40\Delta $$
and
$$E\_6'=\frac{1}{2}(-E\_8+E\_6E\_2) $$
$$E\_6''=\frac{7}{24}(E\_{10}-2E\_8E\_2+E\_6E\_2^2) $$
$$E\_6^{(3)}=\frac{7}{36}(-E\_4^3+3E\_{10}E\_2-3E\_8E\_2^2+E\_6E\_2^3)+168\Delta $$
It's a standard fact that the derivative of a modular form is
quasimodular, so it's not surprising that we have polynomials in
$E\_2$. I am surprised about the appearance of the binomial
coefficients though. Is there a deeper reason for their
appearance? Also, I wonder if the/a pattern continues. For
instance, it would be interesting if it happens that there always
is some $\alpha \in \mathbb{Q}$ so that
$$E\_4^{(n)}-\alpha \sum\_{k=0}^{n} (-1)^{k+n}\binom {n}{k}E\_{4+2n-2k}E\_2^{k}$$
is modular (and similarly for $E\_6$). The other direction you
could ask if the pattern extends is for other modular forms
besides $E\_4$ and $E\_6$. I've taken a handful of derivatives of
other Eisenstein series and saw similar results. You don't get
the binomial coefficients though when you take derivatives of
$\Delta$, so maybe at most something general can be said is for
non-cusp forms.



Binomial coefficients and derivatives of modular forms
The constant $\alpha$ in your question can be in fact written explicitly as $(k)\_n/12^n$, where $(a)\_n=\Gamma(a+n)/\Gamma(n)$ is the Pochhammer symbol (shifted factorial) and $k$ denotes the (even) weight of the corresponding Eisenstein series.


Your observation is indeed related to the Rankin--Cohen brakets; see Section 5.2 in [[D. Zagier, Elliptic modular forms and their applications, *The 1-2-3 of modular forms*, Universitext (Springer, Berlin, 2008), pp. 1–-103]](http://dx.doi.org/10.1007/978-3-540-74119-0_1). Preserving the notation $D$ of Zagier's lectures for your differential operator and picking a modular form $f$ of weight $k$, one can show that $D^nf$ transforms under the modular group as
$$
D^nf\biggl(\frac{a\tau+b}{c\tau+d}\biggr)
=\sum\_{r=0}^n\binom{n}{r}\frac{(k+r)\_{n-r}}{(2\pi i)^{n-r}}
c^{n-r}(c\tau+d)^{k+n+r}D^rf(\tau),
$$
by the induction on $n\ge 0$. In addition, the function $E\_2(\tau)$ transforms as
$$
E\_2\biggl(\frac{a\tau+b}{c\tau+d}\biggr)
=\frac{12c(c\tau+d)}{2\pi i}+(c\tau+d)^2E\_2(\tau).
$$
Therefore, it remains to verify that the difference
$$
g\_n=D^nE\_k-\frac{(k) \_ n}{12^n}\sum\_{r=0}^n(-1)^{n-r}\binom{n}{r}E\_{k+2n-2r}E\_2^r
$$
satisfies
$$
g\_n\biggl(\frac{a\tau+b}{c\tau+d}\biggr)=(c\tau+d)^{k+2n}g\_n(\tau).
$$


**Technicalities.**
Indeed, I found the remaining details quite boring, but going through my yesterday
writing I have realised that your expectation *fails* already for $D^5E\_4$ and $D^4E\_6$.
Here is my explanation why.


Because $g\_n(\tau)$ is a $q$-series, so it is invariant under $\tau\mapsto\tau+1$,
we can restrict to verifying the claim under the transformation $\tau\mapsto-1/\tau$
(that is, $a=d=0$, $b=-1$, and $c=1$). Then (we take $s=n-r$ in the above formula)
$$
D^nE\_k(-1/\tau)
=\sum\_{s=0}^n\binom ns\frac{(k+n-s) \_ s}{(2\pi i)^s}\tau^{k+2n-s}D^{n-s}E\_k(\tau)
$$
and
$$
\begin{aligned}
&
\frac{(k) \_ n}{12^n}\sum\_{r=0}^n(-1)^{n-r}\binom nrE\_{k+2n-2r}E\_2^r\bigg|\_{\tau\mapsto-1/\tau}
\cr &\qquad
=\frac{(k) \_ n}{12^n}\sum\_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E\_{k+2n-2r}\biggl(\tau E\_2+\frac{12}{2\pi i}\biggr)^r
\cr &\qquad
=\frac{(k) \_ n}{12^n}\sum\_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E\_{k+2n-2r}\sum\_{s=0}^r\binom rs\frac{12^s}{(2\pi i)^s}\tau^{r-s}E\_2^{r-s}
\cr &\qquad
=\frac{(k) \_ n}{12^n}\sum\_{s=0}^n\binom ns\tau^{r-s}\frac{12^s}{(2\pi i)^s}\tau^{k+2n-s}
\sum\_{r=s}^n(-1)^{n-r}\binom{n-s}{r-s}E\_{k+2n-2r}E\_2^{r-s}.
\end{aligned}
$$
Subtracting the latter from the former we obtain
$$
\begin{aligned}
g\_n(-1/\tau)
&=\sum\_{s=0}^n\binom ns\frac{(k+n-s) \_ s}{(2\pi i)^s}\tau^{k+2n-s}g\_{n-s}(\tau)
\cr
&=\tau^{k+2n}g\_n(\tau)+\sum\_{s=1}^n\binom ns\frac{(k+n-s) \_ s}{(2\pi i)^s}\tau^{k+2n-s}g\_{n-s}(\tau).
\end{aligned}
$$
Therefore, $g\_n(-1/\tau)=\tau^{k+2n}g\_n(\tau)$, hence $g\_n(\tau)$ is a modular form (of weight $k+2n$),
if and only if the additional sum over $s$ vanishes, that is, $g\_{n-s}=0$ for $s=1,\dots,n$. 
The latter however does not happen when $k+2n>12$.


