Question about physics formula?

Wow. You ask a question that requires several semesters of college mathematics to answer.

OK. Here is the bottom line: for a variable x, δx is a infinitely small change in x , often we put it this way: lim (as ε→0) of (x + ε) = x+δx where ε→0 means as ε approaches zero 'smoothly'.

You are right, when the change in x is finite we call it Δx.

The important thing about the differential is that given a function (not *any* function, but many many functions have the right properties to be included in the group of "differentiable" functions that this applies to) y = f(x) then δy/δx = lim(as ε→0) of [f(x + ε)-f(x)] /[(x + ε)-x] which is "rise over run".

This is known as the instantaneous slope of f(x) with respect to x; most importantly it is also known as

the instantaneous rate of change of f(x) w.r.t. x

So, for example if distance = f(time) then δd/δt is the instantaneous rate of cahnge of distance,d, w.r.t. time, t. velocity is = to δd/δt. acelerationg is equal to δ(δd/δt)/δt called the second derivative.

So. the symbol δ is NOT a variable. It symbolizes an operation on the variable(s) following it.

(adding a vanishing ε to it). the pair δx can be considered a variable. Usually we consider δy to be a function of both f(x) and x. Other common operations + ,÷ ,- ,* ,sin(), log(). SOmetimes the operatore is between numbers or variables, but sometimes it is in front of them (sine, tangent, log, exp,)

δ is indeed the lower case delta. It's typically used to show a partial derivative. If you've not had partial differential equations, then you should not be involved with δ and partial differentials.

And, a = δv / δt is NOT typical as it implies partial derivatives when a = dv/dt is the proper definition of acceleration, which is a complete, not partial, derivative. Note that d is used for the complete derivative. δ does not mean infinite change; there is no such thing as infinite change.