One more power before the pattern jumps out. We found that \tfrac{d}{dx}[x^2] = 2x by expanding the difference quotient. Let's do the very same thing for the cubic:

Its graph rises, flattens out as it slips through the origin, then rises again — a gentle S-shape. The slope should be flat at the middle and steep at the ends.

Expand the cube

The only new algebra is the binomial expansion (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3. Put it through the difference quotient:

The x^3 cancels, one h divides out, and as h \to 0 the two terms that still carry an h die away — leaving just 3x^2:

The slope is never negative

Because 3x^2 is a square (times 3), the slope of x^3 is always \ge 0 — the curve only ever rises or, for an instant at x = 0, goes momentarily flat. Slide the point and watch: the tangent tips up steeply on both wings and lies flat as it passes through the origin.

See it together

On the left, the cubic f(x) = x^3. On the right, its derivative f'(x) = 3x^2 — an upward parabola that touches zero only at the origin and is positive everywhere else, matching the always-rising curve.