WHY: Why

Ref. Cal-2

WHAT: What/HOW: How to understand?

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

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = y’, \frac{\Delta{y}}{\mathrm{d}x} = y’ + \epsilon(\mathrm{d}x) \Rightarrow \Delta{y} = y’\mathrm{d}x + \epsilon(\mathrm{d}x)\mathrm{d}x $$

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

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

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

$$ \begin{align*} \Delta{t} = t’\mathrm{d}x + o(\mathrm{d}t) & \Rightarrow \mathrm{d}t = t’\mathrm{d}x = \Delta{t} - o(\mathrm{d}t)\\ \Delta{y} & = y’\mathrm{d}t + o(\mathrm{d}t) \\ [& = y’(\Delta{t} - o(\mathrm{d}t)) + o(\Delta{t} - o(\mathrm{d}t))] \\ [& = y’(t’\mathrm{d}x + o(\mathrm{d}t) - o(\mathrm{d}t)) + o(t’\mathrm{d}x + o(\mathrm{d}t) - o(\mathrm{d}t))] \\ & = y’t’\mathrm{d}x + o(t’\mathrm{d}x) \end{align*} $$

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