Introduction to Derivatives

Ride down a mountain road: some stretches make you squeeze the brakes, others barely need a pedal. The steepness of this road is not one number — it changes all along the way. How do you say exactly how steep it is at this very moment? That is the question derivatives answer.

A straight line has only one slope

A straight line keeps its promise: wherever you stand, one step forward lifts you by the same proportion, so a single slope describes the whole line — see Equation of a Straight Line.

A curve gets no such deal. The parabola $y = x^2$ flattens on the left and steepens on the right, so asking for its slope is meaningless until you first ask: slope at which point?

Average slope first, then close in on the instant

Getting the exact steepness in one jump is hard, so step back first. Pick two points on the curve and connect them with a line — that line is a secant. Its slope is vertical change over horizontal progress:

$\frac{f(b) - f(a)}{b - a}$

That is the average slope between the two points — the same idea as average speed in Speed & Motion: the slope of the chord on a position curve is the average speed over that interval.

Now the magic: hold one point still and slide the other along the curve toward it. As $h$ shrinks, the secant slope settles toward a definite value; when the two points nearly coincide, the secant has become a tangent — hugging the curve, its slope is the curve's true steepness at that point.

InteractiveSecant to Tangent

Point x₀0.5Step h1.5

Secant slope

2.500

Tangent slope

1.000

Drag h toward 0 and the secant hugs the tangent — that's the derivative.

Try it: watch the slope settle down

For $f(x) = x^2$ , take secant slopes from $x = 1$ to a point $h$ away. With $h = 1$ you get $(4-1) \div 1 = 3$ ; with $h = 0.1$ you get $2.1$ ; with $h = 0.01$ you get $2.01$ ... marching toward $2$ . So the steepness of the curve at $x = 1$ is $2$ .

The steepness-at-each-point is itself a function, written $f'(x)$ and read "f prime" — its name is the derivative.

What the prime means

$f'(x)$ is a new function: feed it a position $x$ , and it returns the slope of the curve there. Turning $f$ into $f'$ is like turning an odometer into a speedometer.

Test it on x²

For $f(x) = x^2$ , calculus says $f'(x) = 2x$ . Check it:

At $x = 1$ , the slope is $f'(1) = 2$ ;
At $x = 2$ , the slope is $f'(2) = 4$ .

The parabola really does steepen to the right, climbing from slope 2 to 4 — it fits. Spot-check an average too: between $x = 2$ and $x = 4$ , $\frac{16-4}{4-2} = 6$ , which lands neatly between the tangent slopes 4 and 8 at the ends. Everything agrees.

What derivatives are for

A derivative is a speedometer for change:

Differentiate position with respect to time and you get velocity — the number on a car speedometer is really a derivative of position;
Differentiate population, temperature or sales over time and you learn how fast they rise or fall;
Engineers use it to find maxima and minima: where the slope flips from positive to negative, that is the hilltop.

All of calculus stands on this one move: squeezing an average into an instant.

Check yourself

Quick quiz

1. For f(x) = x² the derivative is f'(x) = 2x. What is the slope of the curve at x = 3?
2. For f(x) = x², what is the slope of the secant joining x = 1 and x = 2?
3. Differentiate the curve of position over time and you get?