# Chain rule proof is a bit unclear. What is epsilion in this proof?

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:

1. 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 $e\Delta{x}$

2. 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?

3. 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.

• Not the most helpful comment, but I'd notice that text anywhere...Stewart's Calculus. You'd be much better off finding the proof from Spivak's Calculus. And consult this MSE post to help. Jan 25, 2018 at 19:59
• The answer to the "Why?" at the end of your question is, the graph of $y=f(x)$ might not be a straight line.
– bof
Jan 25, 2018 at 20:01
• Even if one proof is better or clearer than another, I'd still like to understand this one. Jan 25, 2018 at 20:01
• Can you visually clarify @bof? I understand epsilon to be the difference between the two slopes... how do this lead to that delta y equation? Jan 25, 2018 at 20:02
• @Jwan622 Sure. My comment has more to do with the fact that Stewart does a poor job of motivating his proof, and the tools he provides before that proof are not what they should be perhaps. (I'm actually teaching calculus from Stewart's text...unfortunately.) Spivak does an excellent job motivating the proof and illustrating why "the conventional approach" is flawed. (Stewart alludes to this but gives no examples to illustrate the problem, whereas Spivak gives two.) I'd try to work through Spivak's proof first and then return to Stewart to see if you can make snese of it. Jan 25, 2018 at 20:06

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 $\Delta x\rightarrow 0$.
$$\epsilon(a,\Delta x) = \frac{\Delta y}{\Delta x}-f'(a),$$
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.
• $f(x,y)$ means that $f$ is a function of $x$ and $y$, or in other words, the value of $f$ depends on value of $x$ and $y$. So e(a, delta) means "the value of e depends on the values of a and delta" Jan 25, 2018 at 22:42