How to understand the Chain Rule and its proof?

WHY: Why

Ref. Cal-2

WHAT: What/HOW: How to understand?

First and foremost, I do not like people analyzing derivatives using notion dydx\frac{\mathrm{d}y}{\mathrm{d}x}, but yy’ since I would like to identify Δydx\frac{\Delta{y}}{\mathrm{d}x}:

dydx=y,Δydx=y+ϵ(dx)Δy=ydx+ϵ(dx)dx \frac{\mathrm{d}y}{\mathrm{d}x} = y’, \frac{\Delta{y}}{\mathrm{d}x} = y’ + \epsilon(\mathrm{d}x) \Rightarrow \Delta{y} = y’\mathrm{d}x + \epsilon(\mathrm{d}x)\mathrm{d}x

Why can I break down Δydx\frac{\Delta{y}}{\mathrm{d}x}? It is because I specify dx\mathrm{d}x to be the thing to be limited and the error of Δy\Delta{y}. Notice that it is Δy\Delta{y} but NOT dy\mathrm{d}y. I will later specify what dy\mathrm{d}y is.

Since ϵ(dx)dxdx\epsilon(\mathrm{d}x)\mathrm{d}x \ll \mathrm{d}x as dx0\mathrm{d}x \to 0. Therefore, for simplicity without losing any formality, let o(dx)=ϵ(dx)dxo(\mathrm{d}x) = \epsilon(\mathrm{d}x)\mathrm{d}x, we have Δy=ydx+o(dx)\Delta{y} = y’\mathrm{d}x + o(\mathrm{d}x). And it is time for specifying what dy\mathrm{d}y is. After cancelling the error, dy\mathrm{d}y reveals: dy=ydx=Δyo(dx)\mathrm{d}y = y’\mathrm{d}x = \Delta{y} - o(\mathrm{d}x).

Now let’s proof ΔydtΔtdx=Δydx\frac{\Delta{y}}{\mathrm{d}t} \cdot \frac{\Delta{t}}{\mathrm{d}x} = \frac{\Delta{y}}{\mathrm{d}x}:

Δt=tdx+o(dt)dt=tdx=Δto(dt)Δy=ydt+o(dt)[=y(Δto(dt))+o(Δto(dt))][=y(tdx+o(dt)o(dt))+o(tdx+o(dt)o(dt))]=ytdx+o(tdx) \begin{align*} \Delta{t} = t’\mathrm{d}x + o(\mathrm{d}t) & \Rightarrow \mathrm{d}t = t’\mathrm{d}x = \Delta{t} - o(\mathrm{d}t)\\ \Delta{y} & = y’\mathrm{d}t + o(\mathrm{d}t) \\ [& = y’(\Delta{t} - o(\mathrm{d}t)) + o(\Delta{t} - o(\mathrm{d}t))] \\ [& = y’(t’\mathrm{d}x + o(\mathrm{d}t) - o(\mathrm{d}t)) + o(t’\mathrm{d}x + o(\mathrm{d}t) - o(\mathrm{d}t))] \\ & = y’t’\mathrm{d}x + o(t’\mathrm{d}x) \end{align*}

The equation with [blanket] is for understanding what going on. The key is to understand how dt\mathrm{d}t turn to Δt\Delta{t}.

Built with Hugo
Theme Stack designed by Jimmy