Yes - theoretically, this is possible (on paper).
However, differentiators are inherently unstable if real operational amplifiers are used (loop gain crosses the 0dB line with a phase margin of app. 0 deg).
It is possible to stabilize differentiators - however, in this case their function is disturbed and their use in filters is heavily limited.
More than that, because differentiators are high pass filters they do not suppress noise components as good as integrators. That`s the reason, differentiators are not used for filters and oscillators.
the op-amp is correctly wired up with in an inverting circuit configuration. because of the negative feedback through passive components, the "-" terminal is a virtual ground. the node equations (\$V_2\$ is the voltage at the node where are \$R_1\$, \$R_2\$, \$C_4\$, and \$C_3\$ are connected) are:
$$ \left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right)V_2 - sC_4 V_\text{out} = \frac{1}{R_1} V_\text{in}$$
$$ sC_3 V_2 + \frac{1}{R_5} V_\text{out} = 0 $$
from that, i get
$$ \begin{align}
A(s) \triangleq \frac{V_\text{out}}{V_\text{in}} & = \frac{-\frac{1}{R_1} s C_3}{\frac{1}{R_5}\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right) + (sC_4)(sC_3) } \\
\\
& = \frac{-\frac{1}{R_1 C_4} s}{\frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) + \frac{C_4 + C_3}{R_5 C_3 C_4} s + s^2 } \\
\\
& = \frac{-H \omega_0 s}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2} \\
\end{align} $$
equating the corresponding coefficients...
$$ \omega_0^2 = \frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $$
$$ \frac{\omega_0}{Q} = \frac{C_4 + C_3}{R_5 C_3 C_4} $$
$$ H \omega_0 = \frac{1}{R_1 C_4} $$
i think the intent, in the lecture notes posted in the question is that \$ \omega_0 \triangleq 2 \pi f_\text{m} \$
so let \$ C_3 = C_4 \triangleq C \$ and let \$ k \triangleq \omega_0 C \$.
then $$ \frac{1}{R_1} = H \omega_0 C $$ $$ \frac{1}{R_2} = (2Q - H) \omega_0 C $$ $$ \frac{1}{R_5} = \frac{1}{2Q} \omega_0 C $$.
so plug this in for \$R_1\$, \$R_2\$, and \$R_5\$ and see if equality in the three "corresponding coefficients" equations above is met. if so, the transfer function, as given in the question, is correct.
Best Answer
A filter topology, whether it's a passive one like a pi-section or an active one like the Sallen-Key circuit, is just a way to produce some poles and zeros. Generally, you can tune the circuit values (resistances, capacitances, inductances) to move those poles and zeros around in the s-plane.
A filter design, like Butterworth or Chebychev, is a choice of pole and zero locations that gives a certain performance (Butterworth has maximally flat pass-band, and Chebychev filters minimize the error between the real filter and a boxcar filter, for example).
You can use whatever filter topology you like to implement whichever filter design you like. First determine the desired pole and zero locations from the filter design. Then tune your component values to achieve those locations. However, you might find that some topologies require unreasonably large or small component values for certain filter designs, or be excessively sensitive to errors in the component values, and that kind of issue might motivate you to use a different topology.
As to how they were invented, I suspect the guys who came up with them were simply very very clever.