Q:

no distortion amp circuit, max. (d(Vout)/dt )max< slew rate, why not (d(Vin)/dt)max?

I think slew rate is how fast a circuit can react to a given input. I was asked to find the frequency limit where the output would show no distortion and tried to find it by (d(Vin)/dt)max < slew rate. But my answer wasn't correct because (d(Vout)/dt)max < slew rate, is what I needed to calculate.
It still doesn't make sense for me to analyse the rate of change of the output if you look at the definition of slew rate I gave.

A:

Slew rate is how fast the output can change with respect to time. If you reduce the input signal amplitude (or the gain) you can reproduce a higher frequency with minimal distortion. It's not directly related to the input, only to the desired slew rate (rate of change of voltage) measured at the output (for example, 0.5V/us for the LM324). So with a gain of 1, the amplifier could respond to an input changing at +/-0.5V/us, but at a gain of 100, only to +/-5mV/us at the input. 
There is also a gain-bandwidth product for op-amps. If you have a GBW of 1MHz and a gain of 100, the closed-loop gain will have decreased significantly for a sine wave input of 10kHz. At a gain of 5, it will be 200kHz. This is independent of the signal level (but not the gain)- at least until it becomes slew-rate limited. The formula for where it becomes slew-rate limited is: 
\$f_{max} = \frac  {S_r}{2 \pi V_p} \$
where Sr is the maximum slew rate
Vp is the peak voltage of the sine wave at the output
So, with a slew rate of 0.5V/us the maximum frequency for which a 1V peak sine wave would be reproduced relatively faithfully would be 80kHz. 
Those are a couple simplified non-ideal characteristics. Another is crossover distortion. Even at a few hundred Hz you can see significant crossover distortion in an LM324 output when it's operating as class B. 

