I am reading a proof of the Chain Rule in my text and I see this:
I get the first three lines, but then I have an issue here:
Why does $\epsilon$ (the difference between the difference quotient and the slope at a imply the next equation? Why does $\Delta{y} = f'(a)\Delta{x} + e\Delta{x}$. Isn't the (change in y) just the slope * (change in x)? Why do we need the $e\Delta{x}$
On a high level, what is going on there? Is it because the slope at a * x only gives you part of the change in y? Is $f'(a)\Delta{x}$ an incomplete picture of the change in y? Why?
I also don't understand the significance of this line:
If we define $\epsilon$ to be 0 when $\Delta x$ = 0, then $\epsilon$ becomes a continuous function of Dx.
Formally, $f'(a)=\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}$, where $\Delta y=\Delta y(a,\Delta x)=f(a+\Delta x)-f(a)$. Since equality is only true in the limit, you must allow for some error in any finite approximation. That's why for any fixed $x,\Delta x$ you have $\frac{\Delta y}{\Delta x}=f'(a)+\epsilon$. Here $\epsilon=\epsilon(a,\Delta x)$, and what's true is that $\epsilon\rightarrow 0$ as $\Delta x\rightarrow 0$.
Finally, in its other form:
$$\epsilon(a,\Delta x) = \frac{\Delta y}{\Delta x}-f'(a),$$
which is undefined when $\Delta x=0$. However, since the right side is continuous in $\Delta x$, you can make $\epsilon$ continuous in $\Delta x$ by assigning $\epsilon(a,0):=0$. This underscores the regularity (i.e. "niceness") of the derivative approximation, and gives you a sense of how the limit is acheived.