Expression for the derivative of Eisenstein series $G_2$ I am new to number theory, so I am guessing this is a standard formula. I would be grateful for a reference:
We know that the Eisenstein series $G_2$ is quasimodular of level $SL_2(\mathbb Z)$, so the derivative $G'_2$ is also quasimodular. As such, we can express $G'_2$ as a polynomial in $G_2$ with modular coefficients (of level $SL_2(\mathbb Z)$. Could anyone please tell me the coefficients? 
In other words: we know 
$$G'_2=a_0+a_1G_2+a_2G_2^2$$
where $a_0$, $a_1$ and $a_2$ are modular forms of level $\Gamma(1)=SL_2(\mathbb Z)$. Can you please tell me the expressions of $a_0$, $a_1$ and $a_2$?
 A: According to Ramanujan (and Wikipedia) $qdE_2/dq = (E_2^2-E_4)/12$, where $q=e^{2\pi i\tau}$. It should convert to what you want.
http://en.wikipedia.org/wiki/Eisenstein_series
A: Section 1.3 of Zagier's notes express $G_2'$, $G_4'$, $G_6'$ as explicit elements of $\mathbb{C}[G_2,G_4,G_6]$. In particular, (15) shows that your equation holds with
$$ a_0=\frac{5\pi i}{3}G_4,\qquad a_1=0,\qquad a_2=-4\pi i.$$
A: It's not too bad to check these out by hand, at least for a few low-degree examples such as $E_2$. 
To be fair, we should be clear about normalizations before we begin: I will use the normalizations
$$
E_2 = 1 - 24q - 72q^2 - 96q^3  + \cdots
$$
and
$$
E_4 = 1 + 240q + 2160q^2 + 6720q^3 + \cdots
$$
although you can pick your favourite normalization. In such a case we find that
$$
DE_2 = q\frac{d}{dq}E_2 = -24q - 144q^2 - 288q^2  + \cdots
$$
and that
$$
E_2^2 = 1 - 48q + 432q^2 + \cdots
$$
Since we know that $E_2^2$ and $E_4$ will generate the degree 4 terms, we just need to check the first two coefficients. That is, we know that $DE_2 = \alpha E_2^2 + \beta E_4 = (\alpha + \beta) + (-48\alpha + 240\beta) + \cdots$ which then becomes a straightforward linear algebra problem: that is, we are trying to solve
$$
\alpha + \beta = 0 \\
-48\alpha + 240\beta = -24
$$
i.e. $\beta = -\frac{1}{12}$ and $\alpha = \frac{1}{12}$.
I suppose then to fully answer your question: We would pick $a_0 = -\frac{1}{2}E_4$, $a_1 = 0$, and $a_2 = \frac{1}{12}$.
