Having trouble deducing bandwidth for which differentiation occurs with op amp
Am I correct in thinking that the differentiation occurs from 112.179 hz to 27.295k Hz? 



No 


by differentiation, we generally mean the sinusoidal phase current lags voltage by 90 deg. In this circuit, everything at least an octave below the input breakpoint in the response curve meets criteria, even though you have not plotted the phase response.  
Since the input RC breakpoint T=0.01uF*1k6 = 1.6e-5 or f=10kHz (approx) we can say everything below 10kHZ is a derivative although it needs at least an octave to get close to  -90 deg

This starts at too high a frequency for the plateau response to work


open loop gain @10kHz.= 100 and desired plateau gain =60
with |Zc|=R1=1k6 the impedance feedback ratio,@10kHz = 100k/3k2=31




  And that the max amplification is at 1.7507k Hz. 



No .. maybe you meant 10.75 kHz


the Op Amp is an integrator with -1 slope and the peak gain occurs at the intersection of the +1 differentiator slope just above 10kHz where the impedance ratio increases from 31 towards 60 while open loop gain of 100 drops 50% per octave above 10kHz, so it will be around 11kHz at peak gain




  bandwidth is 27.295k - 112.179 = 27182.821 hz? 



We could call this a Bandpass filter with order +1 diff on low side and -1 on high side, and use the definition of -3dB spread to define BW which looks like +/-2kHz or 4kHz BW on your graph at peak minus -3dB on either side.



  It turns into an inverting amplifier at frequencies greater than 27.295


Since it uses the inverting feedback, the output is always inverting except it has voltage 90 deg phase lead up to center then phase lag by 90 deg. so the net phase is -90 then rapid phase shift from 8 to 12kHz  starting at -90deg to -180 (@11kHZ) to -270 deg for increasing frequency.

p.s. try to follow conventions for orientation on power supplies