Question about Circuit Analysis
Start by analyzing the steady state.  That means all capacitors are open circuits.  The first opamp then just inverts Vinput with respect to ground.  The second then inverts it again, but with a gain of 1/2.  Therefore Voutput = Vinput / 2.

Since both opamps are ideal and have feedback so that they operate in their linear region, the  negative inputs of both will be held at 0 V.

Analyzing the dynamic behavior is more tricky, but you already know what the steady state will be after everything settles.  C2 feeds current proportional to the derivative of Vinput into the node at the negative input of the second amp.  This will add to the steady state signal.  Voutput will therefore have a component that is proportional to -dVinput/dt.

C1 and R2 form a high pass filter, which in the feedback path results in a the second stage being a low pass filter.  Whatever the Voutput would have otherwise been, it will be low pass filtered by a single pole at 1/2&pi;R2C1.  When R is in Ohms and C in Farads, then this expression is in Hz.