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Grant Sanderson · Essence of Calculus

The Paradox of the Derivative

A derivative measures an instantaneous rate of change — but 'change' needs two moments in time. Grant resolves the paradox by making the gap infinitesimally small rather than zero.

The paradox of the derivative | Chapter 2, Essence of calculus17:07

Formula bank

\frac{d s}{d t} = d t \to 0 lim \frac{s ( t + d t ) - s ( t )}{d t}

Limit definition of the derivative6:00

s (t) = t^{3}

Example position function4:10

\frac{d s}{d t} = 3 t^{2}

Its derivative10:00

Derivations

Differentiating s(t) = t³ from first principles

Start from the rise-over-run of a tiny step dt.
$\frac{d s}{d t} = \frac{( t + d t ) ^{3} - t ^{3}}{d t}$
↓ Expand the cube
Binomial expansion of (t + dt)³.
$= \frac{t ^{3} + 3 t ^{2} d t + 3 t d t ^{2} + d t ^{3} - t ^{3}}{d t}$
↓ Cancel the t³ terms
Only the terms containing dt survive.
$= \frac{3 t ^{2} d t + 3 t d t ^{2} + d t ^{3}}{d t}$
↓ Divide every term by dt
Now take the limit as dt → 0.
$= 3 t^{2} + 3 t d t + d t^{2}$
↓ Drop the vanishing terms
The terms with dt disappear, leaving the exact derivative.
$= 3 t^{2}$

1▶ 1:00
The paradox stated
Speed is distance over time, which needs two points in time. Yet a speedometer shows a speed at a single instant. How can there be a rate of change at one moment, when change requires a before and after?
2▶ 4:10
Set up a concrete example
Let a car's position after t seconds be s(t) = t³. We want its velocity at a specific instant, say t = 2, not its average speed over a long stretch.
$s (t) = t^{3}$
- $s$ = distance travelled (m)
- $t$ = time elapsed (s)
3▶ 6:00
Replace the instant with a tiny interval
Instead of asking for change 'at' t, look at the change over a small window dt and divide by dt. This is an honest rate of change — it just happens over a very short time.
$\frac{d s}{d t} = d t \to 0 lim \frac{s ( t + d t ) - s ( t )}{d t}$
- $d t$ = a tiny step in time
- $d s$ = the resulting change in distance
Limit definition of the derivative
4▶ 10:00
Why dt is never actually zero
If you literally set dt = 0 you get 0/0 — nonsense. The trick is to simplify the expression first, THEN ask what it approaches as dt shrinks. The answer, 3t², is the slope the difference quotient is heading toward.
$\frac{d s}{d t} = 3 t^{2}$
💡 Pro tip: Read 'dt → 0' as 'as small as you like, but not zero' — that distinction is the whole resolution to the paradox.

The takeaway

The derivative isn't the slope at a single instant; it's the limit of the slope over a tiny interval as that interval shrinks toward (but never reaches) zero.

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