# Regarding the derivative of the $j$-invariant

Is anyone aware of a formula for the derivative of the $j$-invariant $j(\tau)$ with respect to $\tau$? Here, $\tau$ is in the upper half-plane.

I would image there are probably quite a few formulae for $j'(\tau)$, but they are not well-known. I looked through a few books but I could not find a single formula.

Any help would be greatly appreciated.

Yes, $$\frac{j'(\tau)}{j(\tau)}=-\frac{E_6(\tau)}{E_4(\tau)}$$ (here $$\ '=\frac{d}{2 \pi i d \tau}$$), where $$E_4$$ and $$E_6$$ are Eisenstein series.
Here's a straightforward proof: Use $$j(\tau)=\frac{E_4^3(\tau)}{\eta(\tau)^{24}}$$, known fact that $$\vartheta_{k}(f):=f'-\frac{k}{12}E_2(\tau) f$$ maps modular forms of weight $$k$$ to modular forms of weight $$k+2$$, and $${\rm dim}(M_{14}(SL(2,\mathbb{Z}))=1$$, spanned by $$E_4^2(\tau) E_6(\tau)$$. This allows you to conclude that $$\frac{j'(\tau)}{j(\tau)}$$ is proportional to $$\frac{E_6(\tau)}{E_4(\tau)}$$.
One may recursively prove the formula, for any $$\gamma\in\mathrm{SL}_2(\mathbb R)$$, $$\frac{d^k}{dz^k}(f|_{\gamma,1-k})=\big(\frac{d^k}{dz^k}f\big)|_{\gamma,k+1}$$ where if $$\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$, then $$f_{\gamma,k}(z):=(cz+d)^{-k}f(\gamma z).$$ The formula shows $$\frac{d^k}{dz^k}$$ of a modular function of weight $$1-k$$ is a modular function of weight $$k+1$$. In particular, since $$j$$ is nearly holomorphic of weight $$2$$ (i.e., is holomorphic everywhere except for $$\infty$$), $$j'(z):=\frac1{2\pi i}\frac d{dz}j(z)=q\frac d{dq}j(z)$$ is nearly holomorphic of weight $$2$$.
Moreover, $$j(z)=q^{-1}+O(1)$$, so $$j'(z)=-q^{-1}+O(1)$$. Thus, $$\Delta\cdot j'$$ is a modular form of weight $$12+2$$ (since $$\Delta=\eta^{24}$$ is cuspidal). However, the space of modular forms is the graded algebra $$\mathbb C[E_4,E_6]$$, so $$M_{14}=\mathbb C E_4^2E_6$$. Thus, $$j'(z)\Delta(z)=c E_4^2E_6$$ for some $$c\in\mathbb C$$. Comparing Fourier coefficients tells us $$c=-1$$.