Art Of Electronics - RC Differenciator Circuits and High Pass Filters
Can someone point out where the book finds the required conditon Vout being much less than Vin in section 1.14?

I=C(d/dt(Vin-V))=V/R

S0

I=RC(d/dt(Vin-V))=V

I=RC(d/dt)(Vin)-(d/dt)(RCV)=V

But R and C are defined to be small enough so it means the quantity (d/dt)(RCV) can be considered as approximately  as zero. 
Or the quantity RCV is a  approximately  as zero. 

So

I=RC(d/dt)(Vin)-(d/dt)(RCV)

is approximately equal to

I=RC(d/dt)(Vin)=V

thus

RC(d/dt)(Vin)=V=

C(d/dt)(Vin)=V/R

hence yielding the equation given in section 1.14

of the text book page  shown in photo
This implies the  relation between  Vout &amp; Vin you consider