Ⅶ Logic, Analysis & Beyond · Stage 36 — Limits & Derivatives · 36.3 What a Derivative IsAll lessons →
Stage 36 · Limits & Derivatives

What a Derivative Really Is: Concept and Geometry

One formula for "how fast, right here" — and the tangent line it draws.

Ages 15–19 · Reasoning, one step at a time
The curve f(x) = x² (blue) with its tangent at (1, 1) (green), and faintly the derivative f′(x) = 2x (slate, dashed) — the slope of f, point by point.

In Lesson 36.2 you watched a secant line tip over into a tangent as the gap Δx shrank to nothing. That limiting slope earned a name and a symbol: f′(x₀), the derivative of f at x₀. This lesson pins it down. We write the whole shrinking process as a single formula, work it out by hand for f(x) = x² and watch it land on exactly 2x₀, and read its geometry — the derivative is the slope of the tangent line, so point-slope hands us the tangent's equation. Then comes the leap: collect the slope at every point and f′ stops being one number and becomes a whole new function. Finally a caution — every differentiable curve is continuous, yet a sharp corner like y = |x| is continuous and still has no derivative, because the slope coming from the left disagrees with the slope coming from the right.

36.3.1 The definition of the derivative

The average rate of change of f over the stretch from x₀ to x₀ + Δx is the slope of the secant joining those two points:

average rate = f(x₀ + Δx) − f(x₀)Δx

That is the difference quotient. Lesson 36.2's whole story was: don't stop at one Δx — let it shrink. The number the quotient heads toward, as Δx → 0, is the instant rate of change. We give it a name:

f′(x₀) = limΔx → 0 f(x₀ + Δx) − f(x₀)Δx

Worked example — f(x) = x² by hand

Build the difference quotient and simplify before letting Δx vanish:

(x₀ + Δx)² − x₀²Δx = x₀² + 2x₀Δx + Δx² − x₀²Δx = 2x₀Δx + Δx²Δx = 2x₀ + Δx

The Δx in the bottom cancelled — no more 0/0. Now let Δx → 0: the leftover Δx melts away and we are left with f′(x₀) = 2x₀. So f′(2) = 4, f′(−1) = −2, f′(0) = 0. The slope of x² at any point is exactly twice the x-coordinate.

Notice the trick: we never plugged in Δx = 0 (that would be 0/0). We simplified first, then took the limit. That two-step move — algebra, then limit — is the heart of every derivative computed from the definition.

Try it Shrink Δx and watch the quotient land on 2x₀
For f(x) = x² at the fixed point x₀ = 1, the difference quotient simplifies to 2x₀ + Δx = 2 + Δx. Step Δx down and read the green limit it is closing on.
Δx 2

36.3.2 The slope of the tangent line

Here is the geometry. Pick a point on the curve and zoom in. The more you magnify, the straighter the curve looks around that point — until, very close in, the curve is indistinguishable from a single straight line. That line is the tangent, and its slope is f′(x₀). This is called local linearity: under enough magnification, a smooth curve is its tangent.

Zooming in on f(x) = x² at (1, 1). The tangent has slope f′(1) = 2 (exact, from f′(x) = 2x). The closer you look, the more the blue curve hugs the green line — that steepness is the derivative.
The two faces of f′(x₀)

Analysis: f′(x₀) is the limit of the difference quotient — the instant rate of change. Geometry: f′(x₀) is the slope of the tangent line to y = f(x) at x₀. They are the same number, seen two ways.

36.3.3 Finding the equation of a tangent

Once you know the slope, the line is easy. A tangent passes through the point of contact (x₀, f(x₀)) with slope f′(x₀). Feed those into point-slope form, y − y₀ = m(x − x₀):

y = f′(x₀)·(x − x₀) + f(x₀)

Worked example — tangent to x² at x₀ = 3

Point of contact: (3, f(3)) = (3, 9). Slope: f′(3) = 2·3 = 6. Point-slope:

y = 6(x − 3) + 9 = 6x − 18 + 9 = 6x − 9.

Check the contact point: at x = 3, y = 6·3 − 9 = 9. ✓ The line just kisses the parabola at (3, 9).

Try it Slide the point of contact and read the tangent's equation
Drag x₀ along f(x) = x². The slope is the exact derivative m = f′(x₀) = 2x₀ (from the power rule), and point-slope assembles y = m·x + b for you.
x₀

36.3.4 The derivative is itself a function

So far f′ has been one number at one point. But there is nothing special about that point — every x has its own tangent, its own slope. Collect the rule "x ↦ slope of f at x" and you have built a brand-new function, the derivative function f′(x). For f(x) = x², that rule is f′(x) = 2x — a straight line.

Plotting f and f′ on the same axes makes the link visible. Where f is rising, its slopes are positive, so f′ sits above the x-axis. Where f is falling, f′ is below. Where f turns flat (the bottom of the parabola), f′ crosses zero. The height of the green curve at any x is exactly the steepness of the blue curve there.

Try it Sweep x and check: height of f′ = slope of f
The blue curve is f(x) = x²; the green dashed curve is its derivative f′(x) = 2x. Move the marker: the tangent's steepness on f equals the height of f′ — both equal 2x, computed exactly.
x
A new machine from an old one

Differentiation takes a function f and returns another function f′. Feed f′ to the machine again and you get f″ (the second derivative) — the rate at which the slope itself changes. The derivative is a verb you can repeat.

36.3.5 Differentiability and continuity

If a curve has a tangent at x₀, it certainly can't have a break there — a jump or a hole would make the secants fly apart instead of settling on one line. So differentiable ⟹ continuous: having a derivative is the stronger condition.

The arrow does not reverse. A curve can be perfectly continuous — drawn without lifting the pen — and still have no derivative at a point. The textbook case is a corner: y = |x| at x = 0. Approach the corner from the left and every secant has slope −1; approach from the right and every secant has slope +1. The two one-sided limits disagree, so there is no single slope, no single tangent, and f′(0) does not exist — even though the function is continuous at 0.

Try it The corner of y = |x|: left slope vs right slope
Toggle the side you approach the corner (0, 0) from. From the left the secant slope is −1; from the right it is +1. They never agree, so |x| has no derivative at 0 — yet it is unbroken (continuous).
Watch out

Continuous does not imply differentiable. The corner of |x| is continuous yet has no derivative at 0 — drawing without lifting the pen is not enough; the slope must also settle on one value from both sides.

And the tangent is the limit of secants, not merely "a line that touches at one point." A tangent can cross the curve (the tangent to y = x³ at the origin lies through the curve), while a line that grazes a corner at a single point is not a tangent at all.

What to carry forward

The derivative, in one glance
IdeaWhat it says
Definitionf′(x₀) = limΔx→0 [f(x₀+Δx) − f(x₀)] / Δx
Worked from scratchFor f = x²: quotient = 2x₀ + Δx → f′(x) = 2x
Geometryf′(x₀) = slope of the tangent line at x₀ (local linearity)
Tangent equationy = f′(x₀)·(x − x₀) + f(x₀) (point-slope)
f′ is a functioncollect the slope at every x; f rising ⟺ f′ > 0
Smoothnessdifferentiable ⟹ continuous; a corner (|x|) is the counterexample

Exercises

  1. Use the definition to find f′(x) for f(x) = x², then evaluate f′(−3).
    Show the work
    [(x+Δx)² − x²]/Δx = (2xΔx + Δx²)/Δx = 2x + Δx → f′(x) = 2x as Δx → 0. So f′(−3) = 2·(−3) = −6.
  2. Find the equation of the tangent line to f(x) = x² at the point where x₀ = −1.
    Show the work
    Contact point: (−1, 1). Slope: f′(−1) = 2·(−1) = −2. Point-slope: y = −2(x + 1) + 1 = −2x − 2 + 1 = −2x − 1. Check: at x = −1, y = 2 − 1 = 1. ✓
  3. For f(x) = x², at which x is the tangent horizontal (slope 0)? What is the point?
    Show the work
    Horizontal means f′(x) = 0. Since f′(x) = 2x, set 2x = 0 → x = 0. The point is (0, 0) — the bottom of the parabola, where the tangent is the x-axis.
  4. Use the definition to find f′(x) for the line f(x) = 3x + 5. Does the answer match your expectation?
    Show the work
    [(3(x+Δx)+5) − (3x+5)]/Δx = (3Δx)/Δx = 3, with no Δx left over, so f′(x) = 3 for every x. A line's slope is constant — exactly its derivative.
  5. Explain why f(x) = |x| is continuous at 0 but not differentiable there.
    Show the work
    Continuous: limx→0 |x| = 0 = |0|, so the value equals the limit — no break. Not differentiable: the secant slope from the right is +1 and from the left is −1; the two one-sided limits of the difference quotient disagree, so f′(0) does not exist. A corner has no single tangent.
  6. The derivative of f(x) = x² is f′(x) = 2x. On which interval is f increasing, and how does the sign of f′ tell you?
    Show the work
    f′(x) = 2x is positive exactly when x > 0, so f increases on (0, ∞); f′ is negative for x < 0, so f decreases on (−∞, 0). Where the slope is positive, the curve rises — the height of f′ is the steepness of f.

🎯 Quick check

Six questions to lock it in. Tap the answer you think is right.

§ For teachers and parents

This lesson gives the formal definition behind the derivative — f′(x₀) as the limit of the difference quotient — and its geometry as the slope of the tangent line, building on HSF-IF.B.6 (rate of change over an interval) toward the instantaneous rate. Computing f′ from the definition and assembling the tangent via point-slope draws on HSF-BF (building and interpreting functions). The idea that f′ is itself a function, and the corner of |x| showing that continuity does not imply differentiability, are precisely the conceptual foundations of AP Calculus AB derivatives. A good follow-up question at home: "If the slope is positive everywhere, can the function ever come back down?"

eastmath.com · Stage 36 · 36.3 What a Derivative Is · Reasoning, one step at a time