WHY: Why
Ref. Cal-2
WHAT: What/HOW: How to understand?
First and foremost, I do not like people analyzing derivatives using notion dxdy, but y’ since I would like to identify dxΔy:
dxdy=y’,dxΔy=y’+ϵ(dx)⇒Δy=y’dx+ϵ(dx)dx
Why can I break down dxΔy? It is because I specify dx to be the thing to be limited and the error of Δy. Notice that it is Δy but NOT dy. I will later specify what dy is.
Since ϵ(dx)dx≪dx as dx→0. Therefore, for simplicity without losing any formality, let o(dx)=ϵ(dx)dx, we have Δy=y’dx+o(dx). And it is time for specifying what dy is. After cancelling the error, dy reveals: dy=y’dx=Δy−o(dx).
Now let’s proof dtΔy⋅dxΔt=dxΔy:
Δt=t’dx+o(dt)Δy[[⇒dt=t’dx=Δt−o(dt)=y’dt+o(dt)=y’(Δt−o(dt))+o(Δt−o(dt))]=y’(t’dx+o(dt)−o(dt))+o(t’dx+o(dt)−o(dt))]=y’t’dx+o(t’dx)
The equation with [blanket] is for understanding what going on. The key is to understand how dt turn to Δt.