Here's a use of logarithmic derivatives of a specific modular form.

We need one background fact first. Letting $\mathcal H$ be the upper half-plane, 
if $g \colon \mathcal H \rightarrow \mathbf C$ is holomorphic, bounded 
at $i\infty$, and satisfies
$$
g(\tau+1) = g(\tau), \ \ \ \ g(-1/\tau) = \tau^2g(\tau) + c\tau
$$
for some constant $c$, then one can show $g(\tau) = (\pi i/6)cE_{2}(\tau)$, where $E_2(\tau) := 1 - 24\sum_{n \geq 1}\sigma(n)e^{2\pi in\tau}$ is the "fake" weight 2 modular form, where $\sigma(n) = \sum_{d \mid n} d$. By "fake" I mean $E_{2}(\tau+1) = E_{2}(\tau)$ on $\mathcal H$ but 
 $E_{2}(-1/\tau) \not= \tau^2E_{2}(\tau)$, and instead 
 $E_{2}(-1/\tau) = \tau^2E_{2}(\tau) + (6/\pi i)\tau$. 


We'll use that to prove the famous $q$-product of the modular form $\Delta(\tau)$.


**Theorem**. *For $\tau \in \mathcal H$ with* $q = e^{2\pi i\tau}$
$$
\Delta(q) = q\prod_{n \geq 1}(1-q^n)^{24}.
$$


*Proof*. Since $\Delta(\tau)$ is a weight $12$ modular form, 
the logarithmic derivative $\Delta'(\tau)/\Delta(\tau)$ 
satisfies
$$
g(\tau+1) = g(\tau), \ \ \ \ g(-1/\tau) = \tau^2g(\tau) + 12\tau.
$$
Since $\Delta(\tau)$ is nonvanishing, its logarithmic 
derivative is holomorphic on  $\mathcal H$.  Moreover, 
$\Delta'(\tau)/\Delta(\tau) \rightarrow 2\pi i$ as $\tau \rightarrow i\infty$. Therefore $\Delta'(\tau)/\Delta(\tau) = 2\pi iE_{2}(\tau)$ by the result I described at the start.  


Switching variables from $\tau$ to $q = e^{2\pi i\tau}$, $\Delta'(\tau) = 
2\pi iq\Delta'(q)$, where the $'$ represents 
differentiation with respect to $\tau$ on the left 
and with respect to $q$ on the right. Therefore 
\begin{eqnarray*}
\frac{\Delta'(q)}{\Delta(q)} & = & \frac{1}{q}E_{2}(q) \\
& = & \frac{1}{q}\left(1 - 24\sum_{n \geq 1}\sigma_{1}(n)q^n\right) \\
& = & \frac{1}{q}\left(1 - 24\sum_{n \geq 1}\frac{nq^n}{1-q^n}\right) \\
& = & \frac{1}{q} + 24\sum_{n \geq 1}\frac{-nq^{n-1}}{1-q^n}.
\end{eqnarray*}
This is the logarithmic derivative of 
$q\prod_{n \geq 1}(1-q^n)^{24}$ on $|q| < 1$, where the product 
is holomorphic and nonvanishing.
Functions with equal logarithmic derivatives on $|q| < 1$ are equal up to a nonzero scaling factor, so 
$\Delta(q) = cq\prod_{n \geq 1}(1-q^n)^{24}$.  Comparing 
the coefficient of $q$ in the power series expansion of both sides, $c = 1$.  So we're done.