Proof of The Chain Rule

I’m self-studying Stewart’s calculus and I went back to review the proof of the chain rule, but I keep getting confused about a particular line.

The proof starts as follows:

1. $$\Delta y = f(x+\Delta x) - f(x)$$: this part is clear to me.

2. $$\displaystyle{\lim_{\Delta x \to 0}} [\Delta y/\Delta x] = f’(a)$$: this part is clear to me.

3. let $$\varepsilon$$ be a number such that $$\varepsilon = (\Delta y/\Delta x) - f’(a)$$: not exactly sure why they decided to go in this direction. How does one go about logically arriving at this step?

4. $$\Delta y = \varepsilon \Delta x + \varepsilon f’(a)$$: this is clear to me.

5. If we define $$\varepsilon$$ to be 0 when $$\Delta x = 0$$ then $$\varepsilon$$ becomes a continuous function of $$\Delta x$$: I’m also confused here. How can we define $$\varepsilon$$ to be 0 when $$\Delta x = 0$$? Not sure what that means. Further, if $$\Delta x = 0$$ then wouldn’t $$\varepsilon = 0$$ (no change in the difference quotient) $$- f’(a) = -f’(a)$$?

#3 and #5 are preventing me from proceeding. Can anyone help out?

• Is 4. correct? Simple multiplication should result in $Δy=εΔx+f′(a)Δx$. Oct 5, 2022 at 11:07
• ah yes thats my bad Oct 5, 2022 at 21:57

The definition $$\varepsilon = (\Delta y/\Delta x) - f’(a)$$ does not make sense when $$\Delta x=0$$, since you cannot divide by $$0$$. We need a specific definition of $$\varepsilon$$ for the case where $$\Delta x=0$$.

The choice $$\varepsilon=0$$ when $$\Delta x=0$$ is a possible choice, with the advantage of making $$\varepsilon$$ continuous at $$\Delta x=0$$. Indeed, as you mention in $$2.$$, $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=f'(a)$$ so $$\lim_{\Delta x\to 0}\varepsilon = f'(a)-f'(a)=0$$.

As for why the author of the proof decided to define $$\varepsilon$$ that way, the short answer is "because it works" (to prove what it wants to prove), as the proof probably shows after step $$5.$$. In more details, if $$\varepsilon$$ is defined that way, we can write

$$\Delta y=f'(a)\Delta x + \varepsilon \Delta x$$

(note that your step $$4.$$ is incorrect). Expliciting $$\Delta y$$ and $$\Delta x$$, this means that

$$f(x) = f(a) + f'(a)(x-a) + (x-a)\varepsilon$$

Note that $$f(a)+f'(a)(x-a)$$ is a linear function whose graph is the tangent to the graph of $$y=f(x)$$ at $$P(a,f(a))$$, and $$\lim_{\Delta x\to 0}(x-a)\epsilon=0$$, so when $$x$$ is close to $$0$$ (in other words, $$\Delta x$$ is close to $$0$$), we can write

$$f(x) \simeq f(a)+f'(a)(x-a)$$

This means that the graph of $$y=f(x)$$ can be approximated by its tangent line near $$P$$, which is the original intention when defining the tangent line (as the "best approximation" of a curve near a point).