In Week 2, Eq. 2.3 defined the slope of a tangent line as a limit, and promised that this limit would get a name and a whole chapter of its own. This is that chapter.
4.1 The Derivative at a Point
The derivative of f at a, written f′(a), is
— exactly Eq. 2.3, now with a name. When this limit exists, f is differentiable at a, and f′(a) is both the slope of the tangent line at a and (§4.4) the instantaneous rate of change of f there. An equivalent form, replacing a+h with x and h with x−a, is sometimes more convenient:
Example 4.1 — The derivative from the definition
Find f′(3) for f(x) = x2.
Step 1 — Set up the difference quotient (Eq. 4.1)
f(3+h) − f(3)h = (3+h)2 − 9h = 9 + 6h + h2 − 9h
Step 2 — Simplify before taking the limit
6h + h2h = 6 + h, valid for h ≠ 0, which is all a limit as h→0 ever needs.
Step 3 — Take the limit
f′(3) = limh→0 (6+h) = 6
4.2 The Derivative as a Function
Replacing the fixed point a in Eq. 4.1 with a variable x turns the derivative itself into a function:
f′(x) assigns to each x the slope of f's graph there. Its domain may be smaller than f's own domain — a function can fail to be differentiable at points where it is perfectly well-defined (§4.3).
4.3 Differentiability Implies Continuity
If f is differentiable at a, then f is continuous at a. The converse is false — continuity does not guarantee differentiability.
- A corner. f(x) = |x| is continuous at 0, but the left-hand and right-hand difference quotients give −1 and +1 respectively — no single tangent slope exists.
- A vertical tangent. f(x) = x1/3 has a vertical tangent at 0; the difference quotient grows without bound.
- A cusp. f(x) = x2/3 comes to a sharp point at 0 where the one-sided slopes are +∞ and −∞.
4.4 The Derivative as a Rate of Change
Recall from Week 2 §2.2 that Δy/Δx is the average rate of change over an interval. Letting the interval shrink to a point, as in §4.1, upgrades this to the instantaneous rate of change:
In motion problems, if s(t) is position, s′(t) is velocity: the instantaneous rate at which position changes. This single idea — derivative equals instantaneous rate — is what makes the derivative useful outside mathematics, in physics, economics, biology and engineering alike.
4.5 The Constant and Power Rules
The Power Rule holds for every real exponent — positive, negative, or fractional — not just positive integers, though this course proves it directly from Eq. 4.3 only for positive integers (using the binomial expansion) and small integer/fractional cases.
Example 4.2 — Power Rule with a negative and a fractional exponent
Differentiate (a) f(x) = 1/x3, (b) g(x) = √x.
Step 1 — Rewrite each as a power of x
f(x) = x−3; g(x) = x1/2
Step 2 — Apply the Power Rule
f′(x) = −3x−4 = −3/x4; g′(x) = (1/2)x−1/2 = 1/(2√x)
4.6 The Sum, Difference and Constant Multiple Rules
Both follow directly from the corresponding limit laws in Week 2 §2.6, applied to the difference quotient. Together with the Power Rule, they let you differentiate any polynomial term by term.
Example 4.3 — Differentiating a polynomial
Differentiate f(x) = 4x3 − 7x2 + 2x − 5.
Step 1 — Apply the Power Rule to each term, carrying constants through
f′(x) = 4(3x2) − 7(2x) + 2(1) − 0
Step 2 — Simplify
f′(x) = 12x2 − 14x + 2
4.7 The Product Rule
(fg)′ is not f′g′. Test it on f(x) = g(x) = x: the wrong rule would give 1·1 = 1, but the correct derivative of x·x = x2 is 2x, which only Eq. 4.5 produces: 1·x + x·1 = 2x.
Example 4.4 — The Product Rule
Differentiate y = (x2+1)(3x−2).
Step 1 — Identify the two factors and their derivatives
f = x2+1, f′ = 2x; g = 3x−2, g′ = 3
Step 2 — Apply Eq. 4.5
y′ = 2x(3x−2) + (x2+1)(3)
Step 3 — Expand and simplify
y′ = 6x2−4x + 3x2+3 = 9x2 − 4x + 3
4.8 The Quotient Rule
The numerator's term order matters — unlike the Product Rule, the Quotient Rule is not symmetric in f and g, since division is not commutative. A common memory aid: "low d-high minus high d-low, over low squared," where "low" is the denominator and "high" the numerator.
Example 4.5 — The Quotient Rule
Differentiate y = x2−1x+3.
Step 1 — Identify numerator and denominator with their derivatives
f = x2−1, f′ = 2x; g = x+3, g′ = 1
Step 2 — Apply Eq. 4.6
y′ = 2x(x+3) − (x2−1)(1)(x+3)2
Step 3 — Expand the numerator and simplify
2x2+6x − x2+1 = x2+6x+1
y′ = x2+6x+1(x+3)2
4.9 Derivatives of the Trigonometric Functions
Proved from Eq. 4.3 using the addition formulas of Week 1 (Eq. 1.11–1.12) and two limits, limh→0 (sin h)/h = 1 and limh→0 (cos h−1)/h = 0:
The remaining three follow from the Quotient Rule applied to csc, sec and cot in terms of sin and cos:
The limit limh→0 (sin h)/h = 1 that these formulas rest on is only true when h is measured in radians — exactly the radian-measure warning from Week 1 §1.10 come due.
Example 4.6 — Product Rule with a trig function
Differentiate y = x2 sin x.
Step 1 — Identify the factors
f = x2, f′ = 2x; g = sin x, g′ = cos x
Step 2 — Apply the Product Rule (Eq. 4.5)
y′ = 2x sin x + x2 cos x
4.10 The Chain Rule
For a composite function (Week 1 §1.7), the Chain Rule states:
In Leibniz notation, writing y = f(u) and u = g(x):
— differentiate the "outer" function, leaving the inner function untouched inside it, then multiply by the derivative of the inner function. This is the rule you will use more than any other for the rest of the course.
Example 4.7 — The Chain Rule
Differentiate y = sin(3x2+1).
Step 1 — Identify outer and inner functions
Outer: sin(·); inner: u = 3x2+1, u′ = 6x.
Step 2 — Differentiate the outer function, keep the inner function inside
cos(3x2+1)
Step 3 — Multiply by the derivative of the inner function (Eq. 4.8)
y′ = cos(3x2+1)·6x = 6x cos(3x2+1)
4.11 Worked Capstone Example
Example 4.8 — Combining the Product, Quotient and Chain Rules
Differentiate y = x sin(2x)x2+1.
Step 1 — The outer structure is a quotient (§4.8)
f = x sin(2x), g = x2+1, g′ = 2x.
Step 2 — Differentiate f using the Product and Chain Rules (§4.7, §4.10)
f′ = (1) sin(2x) + x·cos(2x)·2 = sin(2x) + 2x cos(2x)
Step 3 — Assemble with the Quotient Rule (Eq. 4.6)
y′ = [sin(2x)+2x cos(2x)](x2+1) − x sin(2x)(2x)(x2+1)2
4.12 Nomenclature
| Symbol | Meaning |
|---|---|
| f′(x), dy/dx | The derivative of f (or y) with respect to x |
| d/dx[·] | "The derivative with respect to x of..." |
| differentiable | Having a derivative (i.e. the limit in Eq. 4.1 exists) |
| difference quotient | [f(a+h)−f(a)]/h, before the limit is taken |
4.13 Problems
- Derivative from the definition. Use Eq. 4.1 to find f′(2) for f(x) = x2+3x.
- Non-differentiable point. Explain why f(x) = |x−2| is not differentiable at x = 2, even though it is continuous there.
- Power Rule. Differentiate f(x) = 5x4 − 3x−2 + √x.
- Product Rule. Differentiate y = (2x−1)(x3+4).
- Quotient Rule. Differentiate y = (3x+1)/(x2−2).
- Trig derivative. Differentiate y = x cos x − sin x.
- Trig quotient. Differentiate y = tan x using the Quotient Rule applied to sin x/cos x, and check the result matches Eq. 4.7c.
- Chain Rule. Differentiate y = (x2+5)4.
- Chain Rule with trig. Differentiate y = cos(5x−1).
- Chain Rule twice. Differentiate y = sin2(3x) (i.e. [sin(3x)]2).
- Combined. Differentiate y = x2 tan(2x).
- Tangent line. Find the equation of the tangent line to y = x3 − x at x = 1.
- Rate of change. If s(t) = t3 − 6t2 gives position in metres, find the velocity s′(t) and evaluate it at t = 2.
4.14 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §3.1 (tangents and the derivative at a point), §3.2 (the derivative as a function), §3.3 (differentiation rules), §3.4 (the derivative as a rate of change), §3.5 (derivatives of trigonometric functions), §3.6 (the chain rule).
- Pearson MyLab Math — the Week 4 homework set is assigned directly from the sections above; see the Calendar for the due date.