ELI5: slope is "rise over run." Over a wide stretch of road that's just average steepness. But what's the steepness at one exact point? Shrink the run toward zero and watch the average slope settle on a single number — that number is the derivative.
Formally: f′(x) = limit, as h → 0, of [f(x + h) − f(x)] / h. Watch it converge for f(x) = x² at x = 3:
fsq(x)=x2 slopewide=1fsq(3+1)−fsq(3) = 7 run of 1: slope 7 slopethin=10−8fsq(3+10−8)−fsq(3) = 6.0000 run → 0: slope 6 ✓ pass abs(slopethin−6)<0.0001 f′(3) = 2·3 = 6 — the power rule in the flesh The engine also differentiates symbolically, and (on Pro) the Steps button beside the result walks through each rule. The pattern you just watched is the power rule: the derivative of xⁿ is n·xⁿ⁻¹.
diff(x2,x) = 2⋅x the power rule: 2x — click Steps to see why Real-world hook: the derivative of position is velocity; the derivative of velocity is acceleration. In business, the derivative of total cost is marginal cost — the price of making one more unit.
fcu(x)=x3 we'll need the cube function below Try it yourself: define dfc(x), the derivative of x³. (Recall the power rule.) We secretly check it against the true slope of x³ at two points — so no peeking at a formula will save you if it's wrong.
dfc(x)=3⋅x2 ✏️ Your turn: define dfc(x) = the derivative of x³. Green when it matches the real slope at x = 2 and x = 4. ✓ pass abs(dfc(2)−(fcu(2+1e−5)−fcu(2−1e−5))/(2∗1e−5))<1e−2andabs(dfc(4)−(fcu(4+1e−5)−fcu(4−1e−5))/(2∗1e−5))<1e−2 matches the true slope of x³ at two points