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## Section2.1Local Linearity

How do you find the slope of a straight line (in two dimensions)? That's easy: divide the rise by the run, that is, find the ratio of the change in “height” to the corresponding change in “distance” traveled. In other words, find the ratio of the vertical displacement to the horizontal displacement. Using the standard labels $x\text{,}$ $y$ for the horizontal and vertical axes, respectively, the slope $m$ of a straight line is given by

$$m = \frac{\Delta y}{\Delta x}\label{ratio}\tag{2.1.1}$$

where the Greek letter $\Delta$ (“Delta”) can be read as “the change in”. For a straight line, it doesn't matter how large these changes are, so long as they correspond to each other, as shown in <<linslope>>.

This notion of slope as a ratio is central to precalculus. But there is an equivalent formulation, which involves rewriting (2.1.1) in the form

$$\Delta y = m \Delta x\text{.}\label{proportion}\tag{2.1.2}$$

In this form, slope is a proportionality constant relating the rise to the run.

A fundamental assumption of (single-variable) calculus is that all quantities are locally linear, that is, are well-represented by straight lines so long as one “zooms in far enough”. To make this idea mathematically precise requires idealization; we defer a discussion of this important topic until <<ideal>>. For now, it is enough to know that “local linearity” can be made precise – in fact, in several ways.

So what does local linearity tell us about slope? Zoom in at any point on any curve until it looks straight. The slope at that point is then easily determined using as a ratio, using (2.1.1). However, to emphasize that we have zoomed in, we replace $\Delta$ with $d$ throughout. If our curve is described by some function $f\text{,}$ that is, if

$$y = f(x)\text{,}\tag{2.1.3}$$

then we usually also replace $\Delta y$ with $\Delta f\text{,}$ and similarly $dy$ with $df\text{.}$ Thus, (2.1.1) is replaced with

$$m = \frac{df}{dx}\text{.}\label{dratio}\tag{2.1.4}$$

Equivalently, the slope at the given point is, using (2.1.2), the “constant” of proportionality between the rise and the run, representing the “steepness” of the curve at that point. Again, we replace $\Delta$ with $d\text{,}$ thus rewriting (2.1.2) as

$$df = m\,dx\text{.}\label{dproportion}\tag{2.1.5}$$

Now, zoom back out. Clearly, steepness varies with position; the slope is only “constant” on a piece of the curve that is small enough. Nonetheless, the slope represents the local proportionality between rise and run everywhere; it is a function of position along the curve.

This notion of a locally linear slope is called the derivative. Using (2.1.4), the derivative can be written as

\begin{equation*} \frac{df}{dx}\text{,} \end{equation*}

which is known as Leibniz notation. Equivalently, we have the rather obvious-looking statement that

$$df = \frac{df}{dx}\,dx\text{.}\label{dddef}\tag{2.1.6}$$

The tautological appearance of (2.1.6) is in fact a highly desirable feature of Leibniz notation. Don't forget that, by the discussion above, the derivative $\frac{df}{dx}$ is not in general constant, but varies from point to point.

Let's consider some examples. As already discussed, the slope of the straight line given by

$$y = mx + b\tag{2.1.7}$$

is just $m\text{.}$ Reinterpreting this slope as the derivative of the linear function

$$f(x) = mx + b\tag{2.1.8}$$

we write the derivative of $f$ as

$$f'(x) = \frac{df}{dx} = m\tag{2.1.9}$$

where we have introduced the Newton notation $f'(x)$ for the derivative of $f\text{.}$  1 Newton actually used dots, rather than primes; the latter were in fact introduced by Lagrange.

Let's consider a more challenging example, such as $g(x)=x^2\text{.}$ Since $g$ is locally linear, we can treat $g$ as linear so long as we zoom in far enough. Thus, near any point $x\text{,}$ we must have

$$g(x+dx) \approx g(x) + \frac{dg}{dx}\,dx\text{.}\tag{2.1.10}$$

Working out both sides, we obtain

$$x^2 + 2x\,dx + dx^2 \approx x^2 + \frac{dg}{dx}\,dx\tag{2.1.11}$$

which tells us both that we can and should ignore the “$dx^2$” term on the left-hand side (zooming in implies that this term is much smaller than the others), and that

$$\frac{dg}{dx} = 2x\text{.}\tag{2.1.12}$$