Difficulty: Introductory | Prerequisites: Limits (Section 1.3–1.6), basic algebra of polynomials and rational expressions.
Section 2.1 introduces the formal definition of the derivative, the single most important concept in differential calculus. Everything from here forward, including differentiation rules, curve sketching, optimisation and related rates, rests on understanding what the derivative actually measures and how the limit definition produces it. You should already be comfortable evaluating limits (including algebraic simplification to cancel indeterminate 0/0 forms) before working through this material. If limits still feel shaky, revisit Sections 1.3–1.6 first.
The derivative of a function at a point is the limit of a difference quotient as the step size h shrinks to zero. Computing it from the definition means plugging the function into that quotient, simplifying until h cancels, then letting h go to 0. This section drills that process for linear, polynomial, rational and radical functions, and uses the result to write tangent-line equations.
Derivative
The instantaneous rate of change of f(x) at a point x = a. Formally, f'(a) = lim(h→0) [f(a+h) – f(a)] / h, provided the limit exists. Think of it as the slope of the curve at exactly one point, rather than over an interval.
Difference quotient
The expression [f(a+h) – f(a)] / h. It measures the average rate of change of f over the interval from a to a+h. In simple terms, it is the slope of the secant line through two points on the curve that are h units apart.
Limit definition of the derivative (standard form)
f'(a) = lim(h→0) [f(a+h) – f(a)] / h. The step size h shrinks to zero, turning the secant slope into the tangent slope.
Alternative definition of the derivative
f'(a) = lim(x→a) [f(x) – f(a)] / (x – a). Instead of a step size h, the moving point x slides toward a. The two forms are equivalent (set x = a + h).
Tangent line
The line that just touches the curve at a single point and has slope equal to the derivative there. Its equation in point-slope form is y – f(a) = f'(a)(x – a). In simple terms, it is the best straight-line approximation to the curve near that point.
Every problem in this section follows the same three steps:
Form the difference quotient. Substitute a+h into f, subtract f(a), and divide by h.
Simplify until h cancels. Expand, combine like terms, factor. The goal is to remove the h in the denominator so the expression is no longer 0/0.
Take the limit as h → 0. Once h is cancelled, substitute h = 0 into what remains.
The algebra in step 2 changes depending on the type of function. Below are worked strategies for each type that appears in this assignment.
For f(x) = 4 – 5x at x = 1:
f(1+h) = 4 – 5(1+h) = 4 – 5 – 5h = –1 – 5h
f(1) = 4 – 5(1) = –1
Difference quotient: [(–1 – 5h) – (–1)] / h = –5h / h = –5
The h cancels immediately. The quotient is –5 for every nonzero h.
Limit: f'(1) = –5
The derivative of a linear function is its slope, everywhere. For y = 4 – 5x, the slope is –5, so f'(1) equals the slope of the line.
For f(x) = x² – 10 at x = –2:
f(–2+h) = (–2+h)² – 10 = 4 – 4h + h² – 10 = h² – 4h – 6
f(–2) = 4 – 10 = –6
Difference quotient: [(h² – 4h – 6) – (–6)] / h = (h² – 4h) / h = h – 4
Limit: f'(–2) = 0 – 4 = –4
Tangent line at (–2, –6): y – (–6) = –4(x – (–2)), which simplifies to y = –4x – 14.
Alternative definition check: f'(–2) = lim(x→–2) [f(x) – f(–2)] / (x+2) = lim(x→–2) (x² – 10 – (–6)) / (x+2) = lim(x→–2) (x² – 4) / (x+2). Factor the numerator using difference of squares: (x–2)(x+2) / (x+2) = x–2. As x → –2, this gives –4. Same answer.
For f(x) = 5x³ at x = 2:
f(2+h) = 5(2+h)³ = 5(8 + 12h + 6h² + h³) = 40 + 60h + 30h² + 5h³
f(2) = 5(8) = 40
Difference quotient: (60h + 30h² + 5h³) / h = 60 + 30h + 5h²
Limit: f'(2) = 60
The key algebra step with cubics is expanding (a+h)³ = a³ + 3a²h + 3ah² + h³. Every term except the constant carries an h, so h factors out cleanly.
For f(x) = 1/(x+4) at x = –2:
f(–2+h) = 1/((–2+h)+4) = 1/(h+2)
f(–2) = 1/(2) = 1/2
Difference quotient: [1/(h+2) – 1/2] / h
To simplify, combine the fractions in the numerator over a common denominator: [2 – (h+2)] / [2(h+2)] = –h / [2(h+2)]. Divide by h: –1 / [2(h+2)].
Limit: f'(–2) = –1 / [2(0+2)] = –1/4
Tangent line at (–2, 1/2): y – 1/2 = (–1/4)(x+2), which simplifies to y = –x/4.
For f(x) = 2√x – 5 at x = 1:
f(1+h) = 2√(1+h) – 5
f(1) = 2(1) – 5 = –3
Difference quotient: [2√(1+h) – 5 – (–3)] / h = [2√(1+h) – 2] / h
The trick here is to multiply numerator and denominator by the conjugate: 2√(1+h) + 2.
Numerator becomes: 4(1+h) – 4 = 4h
Full expression: 4h / [h(2√(1+h) + 2)] = 4 / (2√(1+h) + 2)
Limit: f'(1) = 4 / (2+2) = 1
Tangent line at (1, –3): y – (–3) = 1(x – 1), so y = x – 4.
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}Standard (h-form) limit definition of the derivative.
f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}Alternative (x-form) limit definition. Set x = a + h and the two are identical.
y - f(a) = f'(a)(x - a)Point-slope form of the tangent line at x = a.
The derivative gives the instantaneous rate of change, which is the basis for velocity in physics (rate of change of position), marginal cost in economics (rate of change of total cost per additional unit), and reaction rates in chemistry. Every time you hear "rate of change" in any discipline, a derivative is working underneath.
Students often forget to distribute the negative sign when subtracting f(a). For instance, with f(x) = x² – 10, forgetting to subtract the full (–6) and writing – 6 instead of – (–6) = + 6 throws off the entire calculation.
Cancelling h before it is properly factored from every term in the numerator. If even one term in the numerator does not contain h, something went wrong in the algebra.
Confusing the difference quotient (which still contains h) with the derivative (which is the limit as h → 0). The quotient [f(a+h) – f(a)] / h with h still present is not the derivative; it is an approximation.
With rational functions, students sometimes try to cancel terms across a sum rather than combining fractions over a common denominator first.
⚠️ You will almost certainly be asked to compute a derivative using the limit definition, not shortcut rules. Know the process cold.
⚠️ The conjugate trick for square roots is a perennial exam favourite. If you see a radical in the difference quotient, multiply top and bottom by the conjugate.
⚠️ Tangent line questions combine the derivative computation with point-slope form. You need both the slope (the derivative) and the point (a, f(a)).
⚠️ The alternative definition (x → a form) often pairs with difference-of-squares or other factoring. Recognise that x² – a² = (x–a)(x+a) lets you cancel the (x–a) denominator.
True or False: The difference quotient [f(a+h) – f(a)] / h is the derivative of f at a. (False – it becomes the derivative only after taking the limit as h → 0.)
Fill in the blank: For f(x) = 3x + 7, the derivative f'(x) = ___. (3, the slope of the line.)
True or False: The alternative definition f'(a) = lim(x→a) [f(x)–f(a)] / (x–a) gives a different value from the standard definition. (False – they are equivalent.)
Fill in the blank: To simplify a difference quotient containing √(a+h), multiply numerator and denominator by the ___. (Conjugate.)
True or False: If f is a linear function, the derivative is the same at every point. (True.)
Q: Use the limit definition to find f'(3) for f(x) = x² + 2x.
A: f(3+h) = (3+h)² + 2(3+h) = 9+6h+h²+6+2h = 15+8h+h². f(3) = 15. Quotient: (8h+h²)/h = 8+h. Limit: f'(3) = 8.
Q: For f(x) = 1/(x+1), simplify [f(0+h) – f(0)] / h.
A: f(h) = 1/(h+1), f(0) = 1. Numerator: 1/(h+1) – 1 = [1 – (h+1)] / (h+1) = –h/(h+1). Divide by h: –1/(h+1). The derivative f'(0) = –1.
Q: Find the equation of the tangent line to f(x) = 3√x at x = 4.
A: f(4) = 6. Derivative by limit definition (conjugate method) gives f'(4) = 3/(2√4) = 3/4. Tangent line: y – 6 = (3/4)(x – 4), or y = (3/4)x + 3.
Q: Using the alternative definition, find f'(5) for f(x) = x² – 1.
A: lim(x→5) (x² – 1 – 24)/(x – 5) = lim(x→5) (x² – 25)/(x – 5) = lim(x→5) (x–5)(x+5)/(x–5) = lim(x→5) (x+5) = 10.
This section connects directly to Section 2.2 (basic differentiation rules), which gives you shortcuts so you no longer need the limit definition for every problem. Understanding the definition here is what makes those shortcuts make sense. It also connects to Section 2.6 (tangent lines and linear approximation), where the tangent line you compute here becomes a tool for approximating function values near a point.
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