look your question is not at all 'bakwaas'............this is a question that every person might have with application of calculus in physics.......................

now let us come to your doubt. i feel that you might  have come accross this concept while deriving an expression for the potential at a point....your expression ie. dw=f*(-dx) is absolutely right. but the way uve interpreted it is wrong. here the small work dw is the work done in moving a charge q(which is not a small charge) through a small distance dx. here as q is finite the force experienced by q at any x(x!=0) is also finite. here as your force experienced is varying with x , we are integrating ie. calculating the work done in movin the charge by a small distance dx such that force f is almost same at both the extremities of dx which is nothing but the force experienced at x.........

and then all these small works are added(or integrated) to obtain the final work........

ur 2nd question can be answered by simple maths..........

E is nothing but a ratio of the changes in voltage per change in x agreed?

now however small i make both these dimensions(changes) their ratio is always going to be a finite number(what i want to say is eg 2=6/3=0.000000000000000004/0.000000000000000002) hence it is not dE but simply E................

i hope u understood this ...............

as far as your third Q. goes bring a pt charge from 00 to a pt(00=inf) increses the potential of a pt.as per defn. potential of a pt it is the work done in bringing a UNIT positive charge form 00 to  that pt. But potential energy in a charge at a pt is the -ve of the work done in bringing THAT charge from 00 to that pt

= q*(potential at that pt)  (q being the charge)

hope that did the job.................................

So powerpuff girl, please do rate me if you found my expln satisfactory ar correct me if i were wrong.......................