Voltage output waveform of a differentiating amplifier
Driving an IDEAL differentiating amplifier with a triangle would result in an IDEAL squarewave. But this is pure mathematics. There are no ideal circuits, in general.

In your case, we have 

(1) a real (non-ideal) opamp with a finite and frequency-dependent gain (lowpass characteristic), and 

(2) a resistor R1 which disturbs the differentiating properties , but which is necessary for stability reasons.

Therefore, we cannot expect a squarewave function. What we see is the typical output of a highpass-lowpass combination with finite rise and fall times (highpass caused by external feedback elements and lowpass property of the opamp).

Hence, the output is as expected and, therefore, correct.