Differentiator cut-off frequency
Since you are talking about a non-ideal case, then you can no longer use the ideal transfer function. What you have shown has this transfer function:
$$\begin{align}
Z_f&amp;=\frac{1}{\frac{1}{R_f}+sC_f} \\
Z_1&amp;=R_1+\frac{1}{sC_1} \\
H(s)&amp;=\frac{Z_f}{Z_1} \\
H(s)&amp;=\frac{1}{\left(R_1+\frac{1}{sC_1}\right)\left(R_f+\frac{1}{sC_f}\right)} \\
H(s)&amp;=\frac{\frac{s}{R_1C_f}}{s^2+\frac{R_fC_f+R_1C_1}{R_fR_1C_fC_1}s+\frac{1}{R_fR_1C_fC_1}}
\end{align}$$
So you can see that it's a bandpass now, which matches the frequency plot in your second picture. Even if you have this transfer function, you cannot talk about a corner frequency for a differentiator, since, if there is a pole, it's no longer a differentiator, it's a highpass, at least. And things get a bit more complicated due to the internal pole of the opamp, parasitic indictances of the elements, themselves, but for a general (and, usually, whithin the necesssary bandwidth) case, this will do, and you can call it a differentiator.