When the input waveform is significantly faster than the amplifier stage, the leading and trailing edges will no longer be vertical, because the amplifying circuit has a limited bandwidth. It is very easy to perform a square wave test and end up with an entirely wrong answer if you're not careful. Much of the brouhaha that developed regarding TIM (transient intermodulation distortion) and/or SID (slew induced distortion) were due to the very fast risetime of the test signal. When testing any audio device, you must be aware of the simple fact that music does not contain very fast risetime signals, and most media (vinyl, CD, etc.) are actually not very demanding. This is because the amplitude of the musical harmonics is reduced by at least 6dB/octave from no higher than 2kHz or so. This means that the actual level at 20kHz will typically be 20dB lower than the level at midrange frequencies.
Therefore, an amplifier that can provide ±35V peaks will only be required to provide around ±3.5V peaks at 20kHz when operating just below full power with music as the input. This dramatically changes the required slew rate, but it's very common (and advisable) to ensure that an amplifier can reproduce no less than 50% output voltage at 20kHz to ensure an acceptable safety margin. TIM may have been discredited (along with its siblings), but it doesn't make any sense to limit an amplifier if it's not necessary. It also doesn't make sense to go to a great deal of additional effort to design an amplifier that can reproduce full power at 100kHz (or even 20kHz), because it will never be needed.
This is a great start! But I would go further:
Similar to your example, assume a ±35 V waveform with a 6 dB/octave sloping amplitude spectrum. The 6 dB/octave slope gives us that the amplitudes of the harmonics are simply A(n) = A1/n, where n is the harmonic and A1 is the amplitude of the fundamental.
It also follows that a 6 dB/octave slope results in the fundamental and all the harmonics having exactly the same slew rate, 2piA1*f, where f is the fundamental.
So, if I have up to 10th harmonic, then the slew rate requirement for the sum of perfectly phase-correlated sinusoids at zero crossing, at least initially, is 10× the slew rate of any of the harmonics or of the fundamental.
Let's compute the slew rate. A = sum (A1/n) for n = 1..10, so A ≈ A1 * 2.9.
Therefore, A1 = 35 V / 2.9 ≈ 12 V. And the slew rate S1 ≈ 2pi12*2000 = 150 kV/s = 0.15 V/µs.
So the slew rate for the waveform at the initial zero crossing is 1.5 V/µs - again, the required slew rate is 10× that of the 20 kHz harmonic alone.
And, interestingly, if I wanted to test my amp and drive it to exercise the same rate with just a 20 kHz sinusoid, it would need to be ±12 V, it's not quite the initial ±35 V, but it's not small either (it is even more for a waveform with 6db/octave slope power spectrum, but it's a bit more involved, so I won't share it as I am not sure I did it correctly).
So, the "no less than 50% output voltage at 20kHz" rule of thumb clearly holds.
I’ve checked my math a couple of times, but if I messed up anywhere - please don't be mad
Takeaway
: even moderate high-frequency content can demand an order-of-magnitude more slew rate when phase-aligned with lower harmonics.