The derivative measures how fast a function changes. With it we write the equation of a tangent line, find where a function increases or decreases, its extrema and its largest and smallest values, and compute velocity and acceleration. Exams require applying the derivative table and rules without mistakes.
§1. Definition and meaning
If the argument changes by Δx, the function changes by Δf = f(x₀ + Δx) − f(x₀). Δf/Δx is the average rate of change. Its limit as Δx → 0 is the derivative:
f′(x₀) = lim(Δx→0) (f(x₀ + Δx) − f(x₀))/Δx
Geometric meaning:f′(x₀) is the slope of the tangent to the graph at x₀: f′(x₀) = tg α.
Physical meaning: if the distance is s(t), the velocity is v(t) = s′(t) and the acceleration a(t) = v′(t) = s″(t).
Worked example
A point moves by the law s(t) = t³ − 2t² + 3 (m). Find its velocity and acceleration at t = 2 s.
(eˣ)′ = eˣ, (aˣ)′ = aˣ ln a, (ln x)′ = 1/x, (logₐx)′ = 1/(x ln a)
Rules (u, v are functions, C is a number):
(u ± v)′ = u′ ± v′, (Cu)′ = Cu′
(uv)′ = u′v + uv′, (u/v)′ = (u′v − uv′)/v²
Worked example
Differentiate f(x) = (x + 1)/(x − 1).
u = x + 1, v = x − 1, u′ = v′ = 1.
f′ = ((x − 1) − (x + 1))/(x − 1)² = −2/(x − 1)².
§3. The chain rule
If a function sits inside another function, "the outer derivative is multiplied by the inner derivative":
(f(g(x)))′ = f′(g(x))·g′(x)
Samples:
((3x − 1)⁵)′ = 5(3x − 1)⁴·3 = 15(3x − 1)⁴;
(sin²x)′ = 2sin x·cos x = sin 2x;
(e²ˣ)′ = 2e²ˣ;
(ln(x² + 1))′ = 2x/(x² + 1);
(√(1 − x²))′ = −2x/(2√(1 − x²)) = −x/√(1 − x²).
§4. Equation of the tangent line
The tangent passes through (x₀; f(x₀)) and has slope f′(x₀):
y = f(x₀) + f′(x₀)(x − x₀)
If the tangent must be parallel to a line y = kx + b, find x₀ from f′(x₀) = k.
Worked example
Write the tangent to f(x) = x² − 3x + 1 at x₀ = 2.
f(2) = 4 − 6 + 1 = −1.
f′(x) = 2x − 3, f′(2) = 1.
y = −1 + 1·(x − 2) = x − 3.
Worked example
At which point of the graph of f(x) = x² − 3x is the tangent parallel to y = 5x + 1?
f′(x) = 2x − 3 = 5 → x₀ = 4.
f(4) = 16 − 12 = 4. Point (4; 4).
§5. Analysing a function: monotonicity and extrema
f′ > 0 — increasing; f′ < 0 — decreasing
Critical points are where f′ = 0 or f′ does not exist.
If f′ changes sign from + to − — maximum; from − to + — minimum. No sign change — no extremum.
With the second derivative: f′(x₀) = 0 and f″(x₀) > 0 — minimum, f″(x₀) < 0 — maximum. For example, for f(t) = t⁴ − 2t² + 1: f′(1) = 0, f″(1) = 8 > 0, so t = 1 is a minimum.
Largest/smallest value on an interval: compare the values at the critical points inside the interval and at its endpoints.
Worked example
Analyse f(x) = x³ − 3x² + 2 and find its largest and smallest values on [−1; 3].
§6. Physical meaning: velocity, acceleration and tangents with a given slope
The derivative is a rate of change. If a point moves along a line and its position at time t is s(t), the first derivative gives the instantaneous velocity and the second derivative gives the acceleration:
v(t) = s′(t), a(t) = v′(t) = s″(t)
v(t) > 0 — the point moves in the positive direction, v(t) < 0 — it moves back, v(t) = 0 — it is momentarily at rest (the direction may change).
Distance travelled and displacement are different: if the point turns back, the distance is the sum of the absolute displacements on each piece.
The idea goes beyond physics: if V(t) is the volume of water in a tank, V′(t) is the filling rate; if C(x) is the cost of x items, C′(x) is the approximate cost of one more item.
A typical question on the geometric meaning is a tangent parallel to a given line. Parallel lines have equal gradients, so the point of contact comes from f′(x₀) = k, and then the tangent formula from §4 is applied. For example, the tangent to y = x² − 3x parallel to y = x + 5: 2x − 3 = 1 → x₀ = 2, f(2) = −2, tangent y = x − 4.
Worked example
A point moves according to s(t) = t³ − 6t² + 9t (m). When does it stop, what is its acceleration at those moments, and what distance does it cover during [0; 3]?
v(t) = s′(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3); v = 0 at t = 1 and t = 3 s.