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 $2$.