Deriving $F = ma$ - Newton's Second Law of Motion

1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo.

2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $ 4v $ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. Then

$$ \Delta v = v_2 - v_1 \,,$$

and what will $\Delta(4v)$ be? Why it will be

$$ \Delta(4v) = 4v_2 - 4v_1 = 4(v_2 - v_1) = 4\Delta v \,.$$

So the constant comes out the front, because it doesn't change between the start and end points that you're considering. Hence it can be factored out.

3) It's important that I point out that

$$ \frac{\Delta p}{\Delta t} = m \frac{\Delta v}{\Delta t} $$

is not Newton's second law. It's just a relationship between the rate of change of momentum and acceleration, and can be derived straight from the definitions of these things. Newton's second law cannot be derived, and is a statement of real physical content --- hence it is called a law. Newton's law can be written as either

$$ F = m \frac{\Delta v}{\Delta t} \,, $$

or

$$ F = \frac{\Delta p}{\Delta t} \,,$$

whichever you prefer. You can show these two are equivalent using the argument in your textbook (although in fact they're not quite equivalent, as I established above --- the first equation only holds when mass is constant; the second is more general, and more universally true).

The point is that Newton's second law tells us how the acceleration or rate of change of momentum of an object is related to the force acting on it.

Hope this helps!

1. Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$.

2. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a fixed and constant mass, $m$. This means that $dm=0$ at all times by definition. We have, using the product rule (which only holds for infinitessimal changes): $$\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ Can you guess what happens to the first term on the right hand side of the equal sign?

You are right, there is a $\Delta$ missing in front of the $t$.

$\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps.

The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like that one does appear in the complete analysis of systems of variable mass (like a rocket with its propellent being exhausted from the rear). It is also sometimes called Newton' second law in incorrect analyses of the rocket problem.