Derivative of y = x^3

Here is a fact that catches nearly everyone off guard. Take a cube-shaped box — say 1 metre along each edge — and double its side. You do not get double the volume. You get eight times the volume: 2^3 = 8. Triple the side and the volume explodes by a factor of 27. Volume is ferociously sensitive to side length, and it gets more sensitive the bigger the cube already is.

"How sensitive, exactly?" is a derivative question. If V = s^3, then \tfrac{dV}{ds} tells you how fast volume responds to a nudge in the side — and this page proves that the answer is 3s^2. At a side of 1 m, volume grows at 3 m³ per extra metre of side; at a side of 10 m, it grows at a colossal 300 m³ per metre. Same shape, a hundred times touchier.

There is a second reason to care. You have already found \tfrac{d}{dx}[x^2] = 2x, and before that the derivatives of a constant and of x itself. That's three rungs of a ladder: 0, then 1, then 2x. One more rung — the cube — and a pattern will be staring you in the face.

f(x) = x^3

Its graph is a gentle S: it climbs steeply from the lower left, flattens for a single instant as it slips through the origin, then climbs steeply away to the upper right. So before any algebra, we can already sketch what the slope should do: big, then zero, then big again — and, curiously, never negative. Keep that prediction in mind; the algebra had better agree with it.

The proof, done honestly

The recipe is the same difference quotient as always — step a distance h along, measure the rise, divide, then let h shrink. The only new algebra is expanding a cubed bracket. Do it in two honest stages, no shortcuts: square first, then multiply by one more (x+h):

(x+h)^3 = (x+h)\,(x+h)^2 = (x+h)\,(x^2 + 2xh + h^2) = x^3 + 3x^2h + 3xh^2 + h^3.

Read the four terms as a story about size. x^3 is the old value — it will cancel. 3x^2h carries one factor of the small step h: small. 3xh^2 carries two: tiny. h^3 carries three: microscopic. Now feed the expansion into the difference quotient:

\frac{(x+h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2.

The x^3 cancelled, and one factor of h divided out of every remaining term — that division is what dissolved the \tfrac{0}{0} trap. Now watch who survives the limit. As h \to 0, the terms 3xh and h^2 still carry a factor of h, so they die away to nothing. Only the term that had exactly one h in the expansion — 3x^2h, now shaved down to 3x^2 — makes it through alive:

\frac{d}{dx}\bigl[\,x^3\,\bigr] = \lim_{h \to 0}\left(3x^2 + 3xh + h^2\right) = 3x^2.

Spot-check it with numbers before trusting it. At x = 2 the formula claims a slope of 3 \cdot 2^2 = 12. Take a genuine secant from 2 to 2.1: \tfrac{2.1^3 - 2^3}{0.1} = \tfrac{9.261 - 8}{0.1} = 12.61. Shrink the step to 0.01: \tfrac{8.120601 - 8}{0.01} = 12.0601. The secants are marching straight at 12, exactly as 3x^2 + 3xh + h^2 predicts (put x = 2, h = 0.01 in and you get 12.0601 on the nose). The formula and the arithmetic tell one story.

See the algebra: a cube growing a shell

That binomial expansion isn't just symbol-shuffling — you can see every term of it. Picture a solid cube of side x, and grow each side by a small h. The new material forms a shell, and the shell splits into exactly the pieces of the expansion. Step through:

The three slabs are the whole story. Each covers one face — area x^2 — with thickness h, so together they contribute 3x^2h of new volume. The rods and the corner cube are crumbs: their volumes, 3xh^2 and h^3, carry extra factors of h, so when you divide the added volume by h and let the shell get thin, they vanish — precisely the terms that died in the limit above. The derivative 3x^2 is not a coincidence of algebra: it is the area of the three faces the cube grows through. Geometry and limit agree because they are the same computation seen from two sides.

The touchiness of x^3 runs the living world. An animal's heat production goes roughly with its volume — with x^3 for body size x — but its heat loss goes with its skin area, only x^2. Scale a mouse up to elephant size and the furnace inside grows far faster than the radiator outside: volume's growth rate 3x^2 simply outruns area's 2x. That's why elephants ship enormous ears (extra radiator), why whales can afford to live in freezing water (heat kept in), and why a shrew must eat almost constantly (heat pouring out).

The same squeeze traps cells. A cell feeds through its surface (x^2) but must feed its whole interior (x^3). Grow too big and the inside outstrips the membrane's ability to supply it — so cells divide instead of ballooning. Biologists call this the square–cube law, and it is exactly the comparison of the two derivatives you now know: 2x versus 3x^2.

The slope is never negative

Now interrogate the formula. 3x^2 is three times a square — and a square can't be negative. So the slope of y = x^3 is always \ge 0: the curve only ever rises or, for one single instant at x = 0, lies momentarily flat. It never goes downhill. Notice the symmetry too: 3(-2)^2 = 3(2)^2 = 12, so the slope at x = -2 equals the slope at x = +2 — the S-curve climbs equally hard on both wings.

Slide the point and watch all of this happen: the tangent tips up steeply out on the wings, eases off as you approach the middle, kisses horizontal exactly at the origin, then steepens again — without ever once tilting downward.

Here is a trap that snares even strong students: "the slope is zero" does not mean "the curve turns around."

With y = x^2, the flat spot at x = 0 really was the bottom of a valley — falling on one side, rising on the other. It's tempting to conclude that f'(a) = 0 always marks a peak or a valley. The cubic says no. At x = 0 the slope of x^3 is exactly 3 \cdot 0^2 = 0 — yet the curve is rising just before it and rising just after it. It doesn't turn; it pauses. Like a cyclist cresting a bridge without pedalling backwards, the curve flattens for one instant and carries on up.

So the honest statement is one-directional: a smooth peak or valley forces the slope to be zero there, but a zero slope alone doesn't force a peak or valley. When you later hunt for maxima and minima, x^3 at the origin is the standard counterexample your examiner hopes you've forgotten. Don't forget it.

Two curves, one story

Plot the cubic f(x) = x^3 (solid) together with its derivative f'(x) = 3x^2 (dashed) and read the pair like an instrument panel: the dashed parabola is a continuous record of the solid curve's steepness. Wherever the cubic is steep, the parabola is high; where the cubic flattens through the origin, the parabola dips to touch zero — and only touches, never crosses, because the slope never goes negative.

There's a small poetry in the shapes: differentiating a cubic hands you a parabola, just as differentiating a parabola handed you a straight line, and a straight line handed you a constant. Each derivative is one degree simpler than the curve it came from. For a second telling of the whole computation, watch it narrated:

The ladder so far — can you smell the pattern?

Line up everything you have proved, one power at a time, each by the same difference-quotient recipe:

f(x) f'(x)
c (a constant)0
x1
x^22x
x^33x^2
x^4?
x^n?

Look at what happens from row to row: the exponent jumps down in front as a multiplier, and the power drops by one. x^2 \to 2x^1; x^3 \to 3x^2. Even the first two rows obey it if you squint: x^1 \to 1 \cdot x^0 = 1, and a constant c \cdot x^0 gets multiplied by its exponent 0 and vanishes.

So — before anyone tells you — write down your guess for the two missing rows. What should \tfrac{d}{dx}[x^4] be? And \tfrac{d}{dx}[x^n] for any whole number n? You can even hear why the guess should be right: expanding (x+h)^4 the same honest way begins x^4 + 4x^3h + \cdots, and by now you know that only the single-h term survives the quotient and the limit. Whatever coefficient sits on that term is the derivative. The next page names this pattern — the power rule — and proves your guess holds for every n at once. Arrive with your prediction already written down; there's no better feeling than being told something you had already discovered.

See it explained