The derivative: definition, geometric meaning and rules

In this lesson you will learn

Compute an average rate of change and explain the derivative as its limit and as the slope of the tangent
Differentiate the basic elementary functions and combine them with the sum, product, quotient and chain rules
Write the tangent line at a point, with a given slope, or through an outside point
Find instantaneous velocity and acceleration from a position function

A car's trip computer shows the average speed of a journey, say 60 km/h, while the speedometer shows the speed at this very instant. The derivative is the mathematics of the speedometer: the rate of change at one point. In the CSCA mathematics paper it brings some of the most predictable points: find f′(x₀), write a tangent, find a velocity. This lesson builds the derivative from scratch and learns its rules; the next one, «Applications of the derivative: monotonicity and extrema», puts them to work.

From the average rate of change to the derivative

Definition

Average rate of change

When x changes from x₀ to x₀ + Δx, the function changes by Δy = f(x₀ + Δx) − f(x₀). The ratio Δy/Δx is the average rate of change of f on this interval (平均变化率). Geometrically it is the slope of the secant through the points (x₀, f(x₀)) and (x₀ + Δx, f(x₀ + Δx)); for a position function s(t) it is the average velocity.

Δy/Δx = [f(x₀ + Δx) − f(x₀)] / ΔxΔy/Δx = [f(x₀ + Δx) − f(x₀)] / Δx

where:

Δxthe increment of the argument (Δx ≠ 0; it may be negative)
Δythe increment of the function

Average rate of change on [x₀, x₀ + Δx] = slope of the secant. On an interval [a, b]: [f(b) − f(a)]/(b − a).

Example 1: average rates of change

1) f(x) = x² on [2, 5].
2) A body moves by the law s(t) = 5t² (m). Find its average velocity from t = 1 s to t = 3 s.
3) f(x) = 1/x on [1, 2].

Show solution

1) [f(5) − f(2)]/(5 − 2) = (25 − 4)/3 = 7.
2) Average velocity = [s(3) − s(1)]/(3 − 1) = (45 − 5)/2 = 20 m/s.
3) (1/2 − 1)/(2 − 1) = −1/2: the function decreases, so the secant falls.

Now let Δx shrink. The second point slides along the graph towards the first, and the secant turns until, in the limit, it becomes the tangent line (切线). The average rate of change becomes the instantaneous rate of change (瞬时变化率) — the number on the speedometer.

Interactive

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The parabola y = x² and the secant through (1, 1) and (1 + d, (1 + d)²); its slope is 2 + d. Move d towards 0: the secant turns into the tangent y = 2x − 1, and its slope tends to f′(1) = 2.

Definition

Derivative

The derivative of f at x₀ (导数, dǎoshù) is the limit of the average rate of change as Δx → 0, if this limit exists and is finite. Doing this at every x gives a new function f′(x), the derivative function (导函数); finding it is called differentiation.

f′(x₀) = lim (Δx→0) [f(x₀ + Δx) − f(x₀)] / Δxf′(x₀) = lim (Δx→0) [f(x₀ + Δx) − f(x₀)] / Δx

where:

f′(x₀)the derivative at x₀ = the instantaneous rate of change
lim (Δx→0)the limit as Δx tends to 0 (from both sides)

The definition of the derivative. The name of the step does not matter — h, Δx or 2Δx; what counts is that the same step stands in the argument and in the denominator.

Example 2: the derivative from the definition

1) Find f′(x₀) for f(x) = x².
2) Find f′(2) for f(x) = 1/x.

Show solution

1) Δy = (x₀ + Δx)² − x₀² = 2x₀Δx + (Δx)², so Δy/Δx = 2x₀ + Δx → 2x₀ as Δx → 0. Hence (x²)′ = 2x.
2) Δy = 1/(2 + Δx) − 1/2 = −Δx / [2(2 + Δx)], so Δy/Δx = −1/[2(2 + Δx)] → −1/4.
The table formula (x⁻¹)′ = −x⁻² gives the same: −1/2² = −1/4.

Example 3 (CSCA style): a limit in disguise

f′(1) = 2. What is lim (Δx→0) [f(1 − Δx) − f(1)] / (3Δx)?
A) −6 B) 2/3 C) −2/3 D) 6

Show solution

The step in the argument is −Δx, so rewrite the denominator with the same step: [f(1 − Δx) − f(1)]/(3Δx) = −(1/3) · [f(1 − Δx) − f(1)]/(−Δx) → −(1/3) · f′(1) = −2/3, answer C.
B forgets that the step is −Δx; A and D multiply by 3 instead of dividing.

When is there no derivative? The limit must be the same from both sides. For y = |x| at x = 0 the slope is −1 on the left and +1 on the right: the graph has a corner, and f′(0) does not exist. A function that has a derivative at x₀ is always continuous there (no jump), but a continuous function may still have no derivative — as |x| shows.

Derivatives of the basic functions

Computing a limit every time would be slow, so the derivatives of the basic elementary functions (基本初等函数) are learned once, as a table. All of them follow from the definition: Example 2 already produced (x²)′ = 2x and (1/x)′ = −1/x². The power rule works for every real exponent, so rewrite roots and fractions as powers first.

(C)′ = 0 (xⁿ)′ = n·xⁿ⁻¹ (sin x)′ = cos x (cos x)′ = −sin x

where:

Cany constant (a number such as 5, π, e², ln 3)
nany real exponent: √x = x^(1/2), 1/x = x⁻¹, 1/x² = x⁻²
xin sin and cos the angle is in radians

Power and trigonometric functions. Special cases worth knowing by heart: (√x)′ = 1/(2√x), (1/x)′ = −1/x².

Example 4: powers, roots and trigonometric functions

Find the derivatives: 1) y = x⁵ and y = √x; 2) y = 1/x²; 3) y = x√x; 4) the slope of the tangent to y = sin x at x = π/3.

Show solution

1) (x⁵)′ = 5x⁴; (√x)′ = (x^(1/2))′ = ½x^(−1/2) = 1/(2√x).
2) (x⁻²)′ = −2x⁻³ = −2/x³.
3) x√x = x^(3/2), so y′ = (3/2)x^(1/2) = (3/2)√x.
4) (sin x)′ = cos x, k = cos(π/3) = 1/2.

Why does sine turn into cosine? Near 0, sin x ≈ x (for x = 0.01, sin x = 0.00999983…), so the slope of the sine curve at 0 is 1 = cos 0, and at its top x = π/2 it is 0 = cos(π/2). The same limit, together with the addition formula sin(x + h) = sin x cos h + cos x sin h, gives (sin x)′ = cos x at every point. In a similar way, e is the base whose exponential curve crosses the y-axis with slope exactly 1 — that is why (eˣ)′ = eˣ.

(aˣ)′ = aˣ ln a (eˣ)′ = eˣ (logₐ x)′ = 1/(x ln a) (ln x)′ = 1/x(aˣ)′ = aˣ ln a (eˣ)′ = eˣ (logₐ x)′ = 1/(x ln a) (ln x)′ = 1/x

where:

athe base: a > 0, a ≠ 1
ee ≈ 2.718, the base with ln e = 1, for which the extra factor disappears
xx > 0 in the logarithms

Exponential and logarithmic functions (see «Power and exponential functions» and «Logarithms and logarithmic functions»). eˣ is its own derivative — that is why e is the favourite base of calculus.

Example 5: exponentials, logarithms and constants

1) y = 3ˣ; 2) y = log₂ x; 3) f(x) = eˣ + ln x, find f′(1); 4) y = e² + ln 3.

Show solution

1) y′ = 3ˣ ln 3.
2) y′ = 1/(x ln 2).
3) f′(x) = eˣ + 1/x, f′(1) = e + 1.
4) Both terms are numbers, so y′ = 0 — not 2e and not 1/3.

Rules: sum, product, quotient and composite functions

(u ± v)′ = u′ ± v′ (Cu)′ = Cu′ (uv)′ = u′v + uv′ (u/v)′ = (u′v − uv′)/v²(u ± v)′ = u′ ± v′ (Cu)′ = Cu′ (uv)′ = u′v + uv′ (u/v)′ = (u′v − uv′)/v²

where:

u, vdifferentiable functions of x; v ≠ 0 in the quotient
Ca constant factor

The rules of differentiation (求导法则). The derivative of a product is NOT the product of the derivatives; in the quotient rule the numerator starts with u′v.

Where does the product rule come from? Think of uv as the area of a rectangle with sides u and v. When x grows by Δx, the sides grow by Δu and Δv, and the area gains two strips, vΔu and uΔv, plus a tiny corner ΔuΔv. Divide by Δx and let Δx → 0: the corner disappears and (uv)′ = u′v + uv′. That is why the answer always has two terms.

Example 6: products and quotients

1) y = x² sin x; 2) y = x·eˣ; 3) y = (x − 1)/(x + 1); 4) show that (tan x)′ = 1/cos² x.

Show solution

1) y′ = 2x sin x + x² cos x.
2) y′ = eˣ + xeˣ = (x + 1)eˣ.
3) y′ = [1 · (x + 1) − (x − 1) · 1]/(x + 1)² = 2/(x + 1)².
4) (sin x / cos x)′ = [cos x · cos x − sin x · (−sin x)]/cos² x = (cos² x + sin² x)/cos² x = 1/cos² x.

Example 7 (CSCA style): the product rule at a point

f(x) = x sin x + cos x. What is f′(π/2)?
A) 1 B) 0 C) −1 D) 2

Show solution

f′(x) = (sin x + x cos x) − sin x = x cos x, so f′(π/2) = (π/2) · 0 = 0, answer B.
C comes from (x sin x)′ = cos x (the product rule ignored): cos x − sin x = −1. D uses (cos x)′ = +sin x: 1 + 0 + 1 = 2. A forgets the derivative of cos x altogether.

[f(g(x))]′ = f′(g(x)) · g′(x); [f(ax + b)]′ = a · f′(ax + b)

where:

gthe inner function
fthe outer function; its derivative is taken at the inner value g(x)
a, bnumbers, a ≠ 0

The chain rule for a composite function (复合函数): outer derivative × inner derivative. The Chinese school programme limits composite functions to the form f(ax + b), so expect this case first of all.

The chain rule says that rates multiply. If u = 3x changes 3 times as fast as x, and sin u changes cos u times as fast as u, then sin 3x changes 3cos 3x times as fast as x. So work from the outside in: differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside.

Example 8: the inner derivative

1) y = sin 3x; 2) y = e^(−2x); 3) y = ln(1 − 3x); 4) y = (3x − 1)⁴; 5) y = √(x² + 1).

Show solution

1) y′ = cos 3x · 3 = 3cos 3x.
2) y′ = e^(−2x) · (−2) = −2e^(−2x).
3) y′ = 1/(1 − 3x) · (−3) = −3/(1 − 3x).
4) y′ = 4(3x − 1)³ · 3 = 12(3x − 1)³.
5) The inner function is x² + 1: y′ = 1/(2√(x² + 1)) · 2x = x/√(x² + 1).

Example 9 (CSCA style): f′(1) inside the formula

f(x) = 2f′(1) · ln x − x, where f′(1) is a constant. What is f′(1)?
A) 1 B) −1 C) 0 D) 1/3

Show solution

f′(1) is just a number, so f′(x) = 2f′(1) · (1/x) − 1. Put x = 1: f′(1) = 2f′(1) − 1, hence f′(1) = 1, answer A (so f(x) = 2 ln x − x).
B treats 2f′(1) · ln x as a constant (its derivative is not 0, because ln x changes); C forgets (−x)′ = −1; D moves 2f′(1) to the wrong side: 3f′(1) = 1.

The tangent line and instantaneous velocity

Geometrically, f′(x₀) is the slope (斜率) of the tangent at the point of tangency (切点) (x₀, f(x₀)): k = tan α, where α is the angle of inclination (倾斜角), 0 ≤ α < π. A positive derivative means the tangent rises (α is acute), a negative one that it falls (α is obtuse), zero that it is horizontal. Parallel lines have equal slopes, and perpendicular lines have k₁k₂ = −1 (see «Lines and circles in the coordinate plane»).

y − f(x₀) = f′(x₀) · (x − x₀)

where:

(x₀, f(x₀))the point of tangency — it lies on the curve and on the tangent
f′(x₀)the slope of the tangent, k = tan α

The equation of the tangent line at x₀ (point–slope form).

1
At the point x₀
Find f(x₀) and f′(x₀), then substitute them into y − f(x₀) = f′(x₀)(x − x₀).
2
With a given slope k
For a tangent parallel to a line, k is that line's slope; for a perpendicular one, k = −1/k₁. Find the point(s) of tangency from f′(x₀) = k — there may be several — and continue as in step 1.
3
Through an outside point P
The point of tangency is unknown: call it (t, f(t)), write the tangent in terms of t, substitute the coordinates of P and solve for t. Chinese problems distinguish «在点P处的切线» (the tangent AT P: P is the point of tangency) and «过点P的切线» (a tangent THROUGH P: P may not even lie on the curve).

Example 10: tangents at a point

1) y = x² − 3x + 1 at x = 2.
2) y = √x at x = 4.
3) The angle of inclination of the tangent to y = x³/3 at x = −1.

Show solution

1) f(2) = −1, f′(x) = 2x − 3, f′(2) = 1: y + 1 = x − 2, y = x − 3.
2) f(4) = 2, f′(4) = 1/(2√4) = 1/4: y − 2 = (x − 4)/4, y = x/4 + 1.
3) y′ = x², k = (−1)² = 1 = tan α, so α = π/4 (45°).

Interactive

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The curve y = x³/3 − x and its tangent at x = a; the slope is k = a² − 1. Move a: at a = ±1 the tangent is horizontal (k = 0), for −1 < a < 1 it falls (k < 0), outside this interval it rises.

Example 11 (CSCA style): a tangent parallel to a line

The tangent to the curve y = x² − x is parallel to the line y = 3x + 5. What is the equation of the tangent?
A) y = 3x − 2 B) y = 3x + 5 C) y = 3x + 4 D) y = 3x − 4

Show solution

Parallel means k = 3: y′ = 2x − 1 = 3 ⇒ x = 2, y = 4 − 2 = 2. Tangent: y − 2 = 3(x − 2), y = 3x − 4, answer D.
A takes y(2) = 4 (it forgets −x); B is the given line itself, which cuts the parabola twice; C has a sign slip in the intercept.

Example 12: tangents through an outside point

Find the tangents to y = x² that pass through the point P(0, −1).

Show solution

P is not on the parabola (0² ≠ −1), so let the point of tangency be (t, t²). The tangent: y − t² = 2t(x − t), i.e. y = 2tx − t².
It passes through P: −1 = −t², so t = ±1.
Two tangents: y = 2x − 1 and y = −2x − 1, touching at (1, 1) and (−1, 1).

v(t) = s′(t) a(t) = v′(t)

where:

s(t)the position (coordinate) at time t, m
v(t)the instantaneous velocity (瞬时速度), m/s; its sign shows the direction
a(t)the acceleration, m/s²

The physical meaning of the derivative: velocity is the rate of change of position, acceleration the rate of change of velocity (see «Kinematics: displacement, velocity, acceleration, free fall»).

Example 13: velocity and acceleration

1) s(t) = 2t³ − 4t (m). Find v and a at t = 2 s.
2) A stone is thrown straight up; its height is h(t) = 20t − 5t² (m). When does it reach the top, and how high is it?

Show solution

1) v = 6t² − 4, v(2) = 20 m/s; a = 12t, a(2) = 24 m/s².
2) v = h′(t) = 20 − 10t. At the top the stone stops for an instant: v = 0 ⇒ t = 2 s, h(2) = 40 − 20 = 20 m. Note that v(0) = 20 m/s is the launch speed and a = v′ = −10 m/s² = −g.

How CSCA asks about this

Compute f′(x₀): a product, a quotient or f(ax + b) at a convenient point (0, 1, e, π/2) — the numbers are chosen so that no calculator is needed.
Limits in disguise: lim [f(x₀ + a·h) − f(x₀ + b·h)]/(c·h) = (a − b)/c · f′(x₀).
Tangent lines: at a point; parallel or perpendicular to a given line (solve f′(x₀) = k first); through an outside point (unknown point of tangency t); a line y = kx + b tangent to a curve — find the parameter.
Reading a tangent: if the tangent at x₀ is y = kx + b, then f′(x₀) = k and f(x₀) = kx₀ + b; the angle of inclination satisfies tan α = f′(x₀), 0 ≤ α < π.
Motion: v = s′(t), a = v′(t); the top of a vertical throw is where v = 0; where v < 0 the body moves backwards.
f′(1) inside the formula: treat it as an unknown number, differentiate, then substitute x = 1.
Time-savers (75 s per item): simplify before differentiating; before choosing a line, check that the point of tangency lies on both the curve and the line; when the options differ only in a number, substitute the number instead of deriving a general formula.

Term	中文	Pinyin
derivative	导数	dǎoshù
derivative function	导函数	dǎohánshù
increment	增量	zēngliàng
average rate of change	平均变化率	píngjūn biànhuàlǜ
instantaneous rate of change	瞬时变化率	shùnshí biànhuàlǜ
limit	极限	jíxiàn
secant line	割线	gēxiàn
tangent line	切线	qiēxiàn
point of tangency	切点	qiēdiǎn
slope	斜率	xiélǜ
angle of inclination	倾斜角	qīngxiéjiǎo
instantaneous velocity	瞬时速度	shùnshí sùdù
basic elementary functions	基本初等函数	jīběn chūděng hánshù
rules of differentiation	求导法则	qiúdǎo fǎzé
composite function	复合函数	fùhé hánshù

Key terms in English and Chinese: you may take the test in either language.

Key points

The average rate of change Δy/Δx is the slope of a secant; the derivative f′(x₀) is its limit as Δx → 0.
Geometric meaning: f′(x₀) = k = tan α, the slope of the tangent at (x₀, f(x₀)); physical meaning: v = s′(t), a = v′(t).
Table: (xⁿ)′ = nxⁿ⁻¹, (sin x)′ = cos x, (cos x)′ = −sin x, (aˣ)′ = aˣ ln a, (eˣ)′ = eˣ, (ln x)′ = 1/x, (C)′ = 0.
Rules: (uv)′ = u′v + uv′, (u/v)′ = (u′v − uv′)/v², [f(ax + b)]′ = a · f′(ax + b).
Tangent: y − f(x₀) = f′(x₀)(x − x₀); for a given slope solve f′(x₀) = k first; for a tangent through an outside point, call the point of tangency t.
Disguised limit: lim [f(x₀ + a·h) − f(x₀ + b·h)]/(c·h) = (a − b)/c · f′(x₀).

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What is the geometric meaning of f′(x₀)?

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