The derivative measures how fast a quantity is changing at a single instant, built from the limit of average rates of change over shrinking intervals. It gives the slope of a tangent line, the instantaneous velocity of a moving object, and — through a few algebraic rules — a way to differentiate almost any function built from elementary pieces.
IntuitionFrom average speed to instantaneous speed
A car's speedometer reports how fast it is going right now, not its average speed over the whole trip. If a quantity is described by a function y=f(x), the average rate of change between x0 and x0+h is the slope of the straight line (the secant line) joining the two points (x0,f(x0)) and (x0+h,f(x0+h)): hf(x0+h)−f(x0). This ratio depends on how large the step h is. To capture the rate of change at the single instantx0, shrink the step: let h→0 and watch the secant line pivot around (x0,f(x0)) until it settles into the tangent line — the best straight-line approximation to the curve at that point.
Interactive plot showing a curve, the secant line through two nearby points at x0 and x0+h, and sliders for x0 and h; as h shrinks toward 0 the secant line rotates to become the tangent line at x0.
The secant line through (x0,f(x0)) and (x0+h,f(x0+h)) for a shrinking step h. Drag h toward 0 to watch the secant pivot into the tangent line at x0; its slope is hf(x0+h)−f(x0)→f′(x0).
SchoolThe limit definition of the derivative
Definition: Derivative at a point
Let f be a function defined on an open interval containing x0. The **derivative of f at x0**, written f′(x0), is the limit of the difference quotient as the step shrinks to zero: f′(x0)=limh→0hf(x0+h)−f(x0), equivalently f′(x0)=limx→x0x−x0f(x)−f(x0). When this limit exists (and is finite), f is called differentiable at x0. Other common notations for the derivative include y′, dxdy, and dxdf(x0).
Geometric meaning. Since the tangent line at x0 is the limit of secant lines, its slope is exactly f′(x0). The tangent line therefore has equation y=f(x0)+f′(x0)(x−x0) — the best linear approximation to the graph of f near x0. A horizontal tangent (f′(x0)=0) marks a point where the curve is momentarily flat, such as at a local extremum.
y=f(x0)+f′(x0)(x−x0)
Physical meaning. If s(t) is the position of an object at time t, the average rate of change hs(t0+h)−s(t0) is average velocity over [t0,t0+h], and its limit v(t0)=s′(t0) is the instantaneous velocity at t0. Differentiating again gives the accelerationa(t)=v′(t)=s′′(t), the second derivative of position — the rate at which velocity itself is changing.
If f is differentiable at x0, then f is continuous at x0.
Why is it true?
A curve with a well-defined tangent slope at x0 cannot jump there: write f(x)−f(x0)=x−x0f(x)−f(x0)⋅(x−x0); as x→x0 the first factor tends to the finite number f′(x0) while the second tends to 0, so the product tends to 0 and f(x)→f(x0).
Proof
This is exactly the argument above: the limit of a product is the product of the limits when both exist, and limx→x0x−x0f(x)−f(x0)=f′(x0) is finite by hypothesis while limx→x0(x−x0)=0.
The converse is false: continuity does not imply differentiability. The standard counterexample is f(x)=∣x∣ at x0=0. It is continuous there, but the difference quotient hf(0+h)−f(0)=h∣h∣ equals 1 for h>0 and −1 for h<0, so the left- and right-hand limits disagree and f′(0) does not exist — the graph has a sharp corner, with no single well-defined tangent slope.
If f,g are differentiable at x, then fg is differentiable at x and (fg)′(x)=f′(x)g(x)+f(x)g′(x).
Why is it true?
Picture f(x)g(x) as the area of a rectangle with width f(x) and height g(x). As x moves, both side lengths change at once, so the rectangle's area changes for two reasons simultaneously: the width grows at rate f′(x) while the height stays at g(x), contributing f′(x)g(x); and the height grows at rate g′(x) while the width stays at f(x), contributing f(x)g′(x). A common mistake is to guess (fg)′=f′g′, but that ignores these two separate contributions entirely.
Proof
Write hf(x+h)g(x+h)−f(x)g(x) and insert −f(x+h)g(x)+f(x+h)g(x) in the numerator to split it into f(x+h)⋅hg(x+h)−g(x)+hf(x+h)−f(x)⋅g(x). As h→0, the first difference quotient tends to g′(x) and f(x+h)→f(x) by continuity, while the second tends to f′(x); the sum tends to f(x)g′(x)+f′(x)g(x).
If g is differentiable at x and f is differentiable at g(x), then f∘g is differentiable at x and (f∘g)′(x)=f′(g(x))g′(x).
Why is it true?
Rates of change compose by multiplying: if y=f(u) changes twice as fast as u (that is f′=2), and u=g(x) changes three times as fast as x (that is g′=3), then over a small step in x, u moves 3 times as far, and y then moves 2 times as far as u moved — so y moves 2×3=6 times as far as x did. A very common mistake is to forget the inner derivative g′(x): dxdsin(x2)=cos(x2); the correct derivative is 2xcos(x2).
Proof
Since f is differentiable at g(x), write f(g(x)+k)−f(g(x))=[f′(g(x))+ε(k)]k for k near 0, where ε(k)→0 as k→0. Substituting k=g(x+h)−g(x), which tends to 0 as h→0 by continuity of g, and dividing by h gives hf(g(x+h))−f(g(x))=[f′(g(x))+ε(k)]⋅hg(x+h)−g(x)→f′(g(x))g′(x).
The cubic curve y = x cubed minus 3x, rising from bottom left to a local maximum near x = -1, falling to a local minimum near x = 1, then rising again to the top right; an inflection point is marked at the origin.
y=f(x)=x3−3x, with f′(x)=3x2−3. The marked local maximum and minimum sit exactly where f′(x)=0, i.e. x=±1: the curve rises while f′(x)>0 (outside [−1,1]) and falls while f′(x)<0 (inside [−1,1]). The marked inflection point at x=0 is where f′′(x)=6x changes sign, i.e. where the curve switches from concave down to concave up.
Example: Differentiating a product
Find f′(x) for f(x)=x2sinx.
Solution
Apply the product rule with u(x)=x2 and v(x)=sinx: u′(x)=2x and v′(x)=cosx, so f′(x)=u′(x)v(x)+u(x)v′(x)=2xsinx+x2cosx.
Example
A ball is thrown upward from ground level. Its height in metres is h(t)=20t−5t2. (a) Find the instantaneous velocity v(t)=h′(t). (b) When does the ball reach its maximum height? (c) What is the speed at t=1 s?
Solution
**Step 1 — Differentiate h(t)=20t−5t2 using the power rule.** Each term differentiates separately: (20t)′=20 and (5t2)′=10t, giving v(t)=h′(t)=20−10t.
Step 2 — Find the time of maximum height. At the peak the ball is momentarily at rest, so v(t)=0: 20−10t=0⇒t=2 s. The second derivative h′′(t)=−10<0 confirms this is a maximum.
**Step 3 — Evaluate the speed at t=1 s.** Substituting into the velocity formula: v(1)=20−10(1)=10 m/s upward. The ball is still climbing at t=1 s, consistent with reaching the peak at t=2 s.
AdvancedHigher-order derivatives and where this leads
Everything above rests on the limit of a function, so a solid grasp of limits is the prerequisite for differentiation. Differentiating f′ itself gives the second derivativef′′=(f′)′ (acceleration, curvature), and repeating the process gives f(n), the n-th derivative; a function with continuous derivatives up to order k is called Ck, and C∞ if it has derivatives of every order. Two directions open up from here: reading off a function's shape — where it increases, has extrema, or is concave — directly from the sign of f′ and f′′ (see applications of the derivative), and running the process in reverse: given f′, recovering f is the problem of antiderivatives, the gateway to integral calculus.
ResearchDerivatives at the research frontier
Using the limit definition, what is f′(3) for f(x)=x2?
Why is f(x)=∣x∣ continuous but not differentiable at x=0?
By the chain rule, what is the derivative of y=sin(2x2)?
A particle moves along a line with position s(t)=t3−6t2+9t (meters), t≥0 in seconds. At what time is its acceleration equal to zero?