Is the derivative of a modular function a modular function Fix a positive integer $n$. Let $f:\mathbf{H}\longrightarrow \mathbf{C}$ be a modular function with respect to the group $\Gamma(n)$. Is the derivative $$\frac{df}{d\tau}:\mathbf{H}\longrightarrow \mathbf{C}$$ also a modular function with respect to $\Gamma(n)$?
I think it's clear that $df/d\tau$ is meromorphic on $\mathbf{H}$ and that it is meromorphic at the cusp. I just don't know why it should be modular with respect to $\Gamma(n)$.

Suppose $f(\tau)$ is modular function of weight $m$, i.e. for $\left( \begin{array}{cc} a & c \\ c & d \end{array} \right) \in \Gamma(n)$, $f\left( \frac{a \tau + b}{c \tau + d} \right) = \left(c \tau + d \right)^{m} f( \tau )$. Differentiating this equality:
$$ \begin{eqnarray}
  \frac{\mathrm{d}}{\mathrm{d} \tau}\left( f\left( \frac{a \tau + b}{c \tau + d} \right) \right) &=& \frac{\mathrm{d}}{\mathrm{d} \tau} \left( \left(c \tau + d \right)^{m} f( \tau )  \right) \\
   f^\prime\left( \frac{a \tau + b}{c \tau + d} \right) \frac{\mathrm{d}}{\mathrm{d} \tau}\left( \frac{a \tau + b}{c \tau + d} \right) &=& \left(c \tau + d \right)^{m} f^\prime(\tau)  + m c \left(c \tau + d \right)^{m-1} f(\tau)\\
   f^\prime\left( \frac{a \tau + b}{c \tau + d} \right) \left( \frac{a d - b c}{(c \tau + d)^2} \right) &=& \left(c \tau + d \right)^{m} f^\prime(\tau)  + m c \left(c \tau + d \right)^{m-1} f(\tau)
  \end{eqnarray}  
$$
Even though $ a d - b c = 1$, the resulting equation shows that the derivative is not a modular function of any weight, except $m=0$, in which case $f^\prime(\tau)$ is a modular function of weight $2$.
