Application 1: in the chapter on modular forms in Serre's Course in Arithmetic, he integrates the logarithmic derivative $f'/f$ of a modular form $f$ on ${\rm SL}_2(\mathbf Z)$ around a contour (Argument principle) to get an important identity (Theorem $3$) that is used later (Theorem $4$, Corollary $1$) to work out dimensions of spaces of modular forms on ${\rm SL}_2(\mathbf Z)$. (This is such a standard way to derive those dimensions that I have to ask: how did you see those dimensions worked out if it wasn't by the method used in Serre's book involving logarithmic derivatives?)

Application 2: We can derive the famous $q$-product of the modular form $\Delta(\tau)$ by working with its logarithmic derivative.

We will 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 $E_{2}(-1/\tau) = \tau^2E_{2}(\tau) + (6/\pi i)\tau$.

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(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.

Application 1: in the chapter on modular forms in Serre's Course in Arithmetic, he integrates the logarithmic derivative $f'/f$ of a modular form $f$ on ${\rm SL}_2(\mathbf Z)$ around a contour (Argument principle) to get an important identity (Theorem $3$) that is used later (Theorem $4$, Corollary $1$) to work out dimensions of spaces of modular forms on ${\rm SL}_2(\mathbf Z)$. (This is such a standard way to derive those dimensions that I have to ask: how did you see those dimensions worked out if it wasn't by the method used in Serre's book involving logarithmic derivatives?)

Application 2: We can derive the famous $q$-product of the modular form $\Delta(\tau)$ by working with its logarithmic derivative.

We will 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 $E_{2}(-1/\tau) = \tau^2E_{2}(\tau) + (6/\pi i)\tau$.

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(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.

Remark. The $q$-product for $\Delta$ is usually attributed to Jacobi, but it's not really due to him. Details on that are here.

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.

Application 1: in the chapter on modular forms in Serre's Course in Arithmetic, he integrates the logarithmic derivative $f'/f$ of a modular form $f$ on ${\rm SL}_2(\mathbf Z)$ around a contour (Argument principle) to get an important identity (Theorem $3$) that is used later (Theorem $4$, Corollary $1$) to work out dimensions of spaces of modular forms on ${\rm SL}_2(\mathbf Z)$. (This is such a standard way to derive those dimensions that I have to ask: how did you see those dimensions worked out if it wasn't by the method used in Serre's book involving logarithmic derivatives?)

Application 2: We can derive the famous $q$-product of the modular form $\Delta(\tau)$ by working with its logarithmic derivative.

We will 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 $E_{2}(-1/\tau) = \tau^2E_{2}(\tau) + (6/\pi i)\tau$.

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(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.