Linear approximation and differentials: Visual walkthrough

Before line one, three plain words we will keep using:

Step 1 — Draw the curve and mark two heights

Figure — Linear approximation and differentials

Step 2 — What "slope at $a$ " even means

Step 3 — Turn "close to $f^{'} (a)$ " into an exact equation

Step 4 — Rearrange into "height = line + error"

Step 5 — Why the error dies faster than the run

Recall How fast, exactly? (needs an extra assumption)

The clean rule "error $\approx C (x - a)^{2}$ " is not free — it holds only when $f$ has a second derivative $f^{''}$ (i.e. $f$ is $C^{2}$ ). Then Taylor series gives the next term $\frac{1}{2} f^{''} (a) (x - a)^{2}$ , so the error is $O ((x - a)^{2})$ . For a function that is merely differentiable (only $f^{'}$ exists), all we may claim is the weaker $o (x - a)$ from the formula above — small compared to the run, but not necessarily square-law. Concavity and second derivative is exactly the $f^{''}$ that governs the quadratic case.

Step 6 — The differential: the same rise, at a working point

Step 7 — Every case: where is $x$ , and which way does the curve bend?

The one-picture summary

This single figure stacks the whole argument: the curve $f$ , the tangent line $L$ touching at $(a, f (a))$ , the run $(x - a)$ , the tangent rise $f^{'} (a) (x - a)$ in red, and the tiny leftover error $ε (x) (x - a)$ — small because it is small $\times$ small.

Recall Feynman retelling — the whole walkthrough in plain words

You are standing on a curvy hill at spot $a$ , and you know two things there: how high you are ( $f (a)$ ) and how steep the ground tilts ( $f^{'} (a)$ ). You want to guess your height a few steps away, at $x$ . So you close your eyes and walk in a perfectly straight line in the direction the ground was tilting. After a run of $(x - a)$ steps, straight-walking has lifted you by slope $\times$ run $= f^{'} (a) (x - a)$ . Add that to where you started: $f (a) + f^{'} (a) (x - a)$ — that is your guess, $L (x)$ . The real hill curved a little while you walked straight, so you are off by a small gap. But here is the magic: that gap is "how much the slope drifted" ( $ε$ ) times "how far you walked" ( $x - a$ ) — two small numbers multiplied, so it is tiny, and it shrinks much faster than your walk as you take fewer steps. That is why walking straight a little way is an almost-perfect shortcut.

Recall Rebuild the derivation from memory

Start from the definition of slope at $a$ ::: $f^{'} (a) = lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}$ . Name the gap between secant and tangent slope ::: $ε (x) = \frac{f ( x ) - f ( a )}{x - a} - f^{'} (a)$ , which $\to 0$ . Multiply through by the run $(x - a)$ and add $f (a)$ ::: $f (x) = f (a) + f^{'} (a) (x - a) + ε (x) (x - a)$ . Why the error is negligible near $a$ ::: it is $ε (x) (x - a)$ = small $\times$ small, so error $/ (x - a) = ε (x) \to 0$ — the error is $o (x - a)$ . Rename base point and run for the differential ::: base $= x$ , step $= d x$ , giving $d y = f^{'} (x) d x$ with $Δ y \approx d y$ .

Connections

Derivative as a limit — Step 2 is literally its definition; the proof is unpacking that limit.
Tangent line — the line $L (x)$ we build in Step 4.
Taylor series — the $(x - a)^{2}$ next term (when $f^{''}$ exists) that pins down the Step-5 error exactly.
Newton's method — repeats Step 4 to chase a root.
Error propagation — Step 6's $d y = f^{'} (x) d x$ in the lab.
Concavity and second derivative — decides over- vs under-estimate in Step 7.