Derivative from first principles — difference quotient definition: Visual walkthrough

Step 1 — What "slope" even means (the staircase)

WHAT. Pick any two points on a graph. Draw a horizontal step (how far right you walked) and a vertical step (how far up you rose). The slope is the answer to one question: for every one step right, how many steps up?

WHY. Before we can talk about the steepness of a curve, we must nail the steepness of a straight line, because that is the only thing we truly know how to measure. Everything later is a trick to reuse this.

PICTURE. In the figure, the horizontal butter-coloured bar is the run (change in the input, how far you moved sideways) and the vertical coral bar is the rise (change in the output, how far you moved up). Slope is rise divided by run.

Figure — Derivative from first principles — difference quotient definition

Step 2 — The trouble with a curve (slope is different everywhere)

WHAT. Look at the parabola $y = x^{2}$ . Near the bottom it is nearly flat; on the right it climbs steeply. There is no single slope — the steepness changes at every point.

WHY. The staircase from Step 1 needs two points. But we want the steepness at one exact point. Two points and one point are in conflict — that conflict is the whole reason the derivative exists.

PICTURE. Three short line segments rest against the curve at three places. Each has a different tilt (flat lavender at the bottom, steeper mint, steepest coral). One curve, many slopes.

Step 3 — The cheat: a buddy point a distance $h$ away

WHAT. Keep your point at input $x$ . Invent a second point a tiny distance to the right. Call that distance $h$ (a small positive number). The second point sits at input $x + h$ .

WHY. We manufacture a second point so the staircase works again — but we keep it close, planning to slide it in later. The letter $h$ is just a name for "how far apart the two points are."

PICTURE. The two points are marked. The lower one is $(x, f (x))$ — its height is $f (x)$ . The upper one is $(x + h, f (x + h))$ — its height is $f (x + h)$ . The horizontal gap between them, labelled in butter, is exactly $h$ .

Step 4 — Build the staircase between them (the secant line)

WHAT. Draw the straight line through both points. It cuts across the curve — this is the secant line. Its rise is the height difference $f (x + h) - f (x)$ ; its run is $h$ .

WHY. We reuse Step 1 exactly, now with our two real points. The slope of this cutting line is a real, computable number — it is the average steepness across the little gap. See Secant and Tangent lines and Average vs Instantaneous Rate of Change.

PICTURE. The coral vertical bar is the rise $f (x + h) - f (x)$ ; the butter horizontal bar is the run $h$ ; the lavender line through both points is the secant.

Step 5 — Why we cannot just set $h = 0$ (the $0/0$ wall)

WHAT. We want the two points to be the same point, so we are tempted to plug $h = 0$ straight into the difference quotient. If we do, the run becomes $0$ and the rise becomes $f (x) - f (x) = 0$ , giving $\frac{0}{0}$ .

WHY. $\frac{0}{0}$ is undefined — it is not a number, it is a question mark. See Indeterminate forms 0 over 0. This is the wall that forces us to do algebra before we shrink $h$ .

PICTURE. As $h$ shrinks, both bars (rise and run) collapse toward zero together. The picture shows three shrinking staircases; the bars get tiny, but their ratio (the tilt of the line) stays sensible and settles on one tilt.

Step 6 — Cancel the $h$ for $f (x) = x^{2}$ (the algebra that saves us)

WHAT. Put $f (x) = x^{2}$ into the difference quotient and simplify until the lone $h$ in the bottom is gone.

WHY. Cancelling $h$ turns the dangerous $\frac{0}{0}$ into a plain expression we can safely evaluate at $h = 0$ . This is the "Cancel" in Plug–Subtract–Cancel–Shrink.

\frac{( x + h ) ^{2} f ( x + h ) - x ^{2} f ( x )}{h} = \frac{x ^{2} + 2 x h + h ^{2} expanded - x ^{2}}{h} = \frac{2 x h + h ^{2} x ^{2} cancels}{h} = \frac{h ( 2 x + h )}{h} = 2 x + h

$(x + h)^{2}$ — the buddy height, every $x$ replaced by the block $x + h$ .
$x^{2} + 2 x h + h^{2}$ — expanded so terms can cancel.
$2 x h + h^{2}$ — what survives after the $x^{2}$ terms kill each other.
$h (2 x + h)$ — factoring pulls one $h$ out front, ready to cancel the bottom $h$ .
$2 x + h$ — safe at last: no division by $h$ .

PICTURE. The algebra mirrors geometry: the secant slope $2 x + h$ is the steepness of the cutting line. The bigger $h$ is, the more the extra $+ h$ tilts it away from the true tangent.

Step 7 — Shrink $h \to 0$ : the secant becomes the tangent

WHAT. In $2 x + h$ , let $h$ slide to $0$ . The leftover $h$ vanishes and we are left with $2 x$ .

WHY. As the gap $h$ closes, the buddy point slides into the original point. The cutting line stops cutting and just touches — it becomes the tangent line. Its slope is the derivative.

f^{'} (x) = h \to 0 lim (stays 2 x + \to 0 h) = 2 x

$lim_{h \to 0}$ — "the value the expression heads toward as $h$ closes in" (see Limits — formal definition and one-sided limits).
$2 x$ — the part with no $h$ ; it does not move.
$h$ — shrinks away to nothing.

PICTURE. A sequence of secant lines pivots about the fixed point; as $h \to 0$ they rotate onto the single coral tangent that grazes the parabola. That tangent's slope is $2 x$ .

Step 8 — Every case, including the degenerate one

WHAT. Check that the recipe behaves on all sign regions and on the flat/edge cases.

WHY. The reader must never meet a scenario we skipped. We test: right arm ( $x > 0$ ), the vertex ( $x = 0$ ), and left arm ( $x < 0$ ); plus the degenerate case of a straight line, where the slope should not change at all.

PICTURE. Tangents drawn at $x = - 2, 0, 2$ : downhill, flat, uphill. Below, a straight line $f (x) = 3 x + 5$ whose tangent is itself — same slope everywhere.

Point $x$	$f^{'} (x) = 2 x$	Tangent looks	Meaning
$x = 2$	$+ 4$	uphill	rising steeply
$x = 0$	$0$	flat	vertex, no tilt
$x = - 2$	$- 4$	downhill	falling steeply

Recall Quick self-test

At which $x$ is the tangent to $y = x^{2}$ horizontal? ::: At $x = 0$ , where $f^{'} (x) = 2 x = 0$ . What geometric event happens as $h \to 0$ ? ::: The secant line rotates until it just touches the curve — it becomes the tangent. Why must we cancel $h$ before letting $h \to 0$ ? ::: Setting $h = 0$ first gives $0/0$ , which is undefined; cancelling removes the bad division.

The one-picture summary

Everything on one canvas: fix the point, draw a family of secants for shrinking $h$ (lavender), watch them pivot onto the single tangent (coral). The rise/run bars shrink together but their ratio settles on $2 x$ — the derivative.

Recall Feynman retelling — the whole walkthrough in plain words

You want to know how steep a hill is at the exact spot where you're standing. But "steepness" needs two points — a here and a there — so you can't measure it standing on one spot alone. Clever fix: plant a flag a tiny step $h$ to your right, run a straight rope from you to the flag, and measure that rope's tilt (up-amount over across-amount). That rope is the secant — it cuts across the hill. For $y = x^{2}$ the tilt works out to $2 x + h$ : mostly $2 x$ , plus a little error $h$ from the flag being too far away. Now pull the flag closer and closer. The error $h$ melts to nothing, the rope stops cutting and just kisses the hill — that's the tangent — and its tilt settles on exactly $2 x$ . That settled number is the derivative: at the bottom of the bowl ( $x = 0$ ) it's $0$ (dead flat), on the right it's positive (uphill), on the left it's negative (downhill). And if the "hill" were really a straight ramp, the rope was already lying flat against it from the start, so the answer is just the ramp's constant slope — no shrinking needed.

Connections

Derivative from first principles — difference quotient definition — the parent recipe this page animates.
Secant and Tangent lines — Steps 4 & 7 are this idea in motion.
Limits — formal definition and one-sided limits — the $lim_{h \to 0}$ machinery of Step 7.
Indeterminate forms 0 over 0 — the wall in Step 5 that forces the algebra.
Average vs Instantaneous Rate of Change — secant = average, tangent = instantaneous.
Power Rule — $2 x$ is the $n = 2$ case of $\frac{d}{d x} x^{n} = n x^{n - 1}$ , proven here from scratch.