Difficulty: Introductory | Prerequisites: Limits (Sec. 1.4 – 1.6), Continuity (Sec. 1.8).
The derivative is the central idea of differential calculus. It measures how fast a function changes at a single point, turning the vague idea of "slope of a curve" into something precise. This section defines the derivative using limits and uses it to write equations for tangent lines. You need to be comfortable evaluating limits (including algebraic tricks like rationalising and simplifying difference quotients) before this will feel routine.
The derivative f'(a) is the limit of the difference quotient as h → 0. It gives you the exact slope of the curve at x = a. Once you have that slope and the point, you write the tangent line in point-slope form.
Derivative at a point, f'(a)
The limit f'(a) = lim h→0 [f(a + h) - f(a)] / h, provided this limit exists. In simple terms, it is the instantaneous rate of change of f at x = a, or the slope of the curve at that one point.
Difference quotient
[f(a + h) - f(a)] / h. This is the slope of the secant line through (a, f(a)) and (a + h, f(a + h)). Think of it as the "average rate of change" over a small interval that you shrink to zero.
Tangent line
The line that just touches the curve at a single point and has slope equal to f'(a). It is the best linear approximation to the curve near that point.
Differentiable at a point
A function is differentiable at x = a if f'(a) exists (the limit above is a finite number). In simple terms, the graph is smooth enough at that point to have a single, well-defined tangent line.
Non-differentiable
A function fails to be differentiable at a point if the graph has a corner (sharp turn), a cusp, a vertical tangent, or a discontinuity there.
The process is always the same:
Write out [f(a + h) - f(a)] / h.
Simplify the numerator (expand, cancel, rationalise) until you can cancel the h in the denominator.
Take the limit as h → 0.
Example: f(x) = 3x + 4, find f'(4).
f'(4) = lim h→0 [f(4 + h) - f(4)] / h = lim h→0 [(3(4 + h) + 4) - (16)] / h = lim h→0 [12 + 3h + 4 - 16] / h = lim h→0 3h / h = lim h→0 3 = 3.
The derivative of any linear function f(x) = mx + b is simply m, its slope.
Once you have f'(a), the tangent line at (a, f(a)) is: y - f(a) = f'(a)(x - a).
Example: y = √x at (1, 1).
f'(1) = lim h→0 [√(1 + h) - 1] / h.
Multiply top and bottom by the conjugate: [√(1 + h) - 1][√(1 + h) + 1] / [h(√(1 + h) + 1)] = [(1 + h) - 1] / [h(√(1 + h) + 1)] = h / [h(√(1 + h) + 1)] = 1 / (√(1 + h) + 1).
As h → 0 this becomes 1/(1 + 1) = 1/2.
Tangent line: y - 1 = (1/2)(x - 1), i.e. y = (1/2)x + 1/2.
Example: y = 3/x at (3, 1).
f'(3) = lim h→0 [3/(3 + h) - 1] / h = lim h→0 [(3 - (3 + h)) / (3 + h)] / h = lim h→0 [-h / (3 + h)] / h = lim h→0 -1/(3 + h) = -1/3.
Tangent line: y - 1 = (-1/3)(x - 3), i.e. y = (-1/3)x + 2.
A function can fail to be differentiable at x = a for several reasons:
Corner or cusp: The graph changes direction abruptly. The left-hand and right-hand limits of the difference quotient exist but disagree (corner) or one or both are infinite (cusp). Example: f(x) = |x - 4| has a corner at x = 4.
Vertical tangent: The slope approaches ±∞. The tangent line exists but is vertical. Example: f(x) = (x - 4)^(1/3) at x = 4.
Discontinuity: If f is not continuous at x = a, f cannot be differentiable there. A jump or hole automatically rules out a derivative.
Note: differentiability implies continuity, but continuity does not imply differentiability.
Limit definition of the derivative:
f'(a) = lim h→0 [f(a + h) - f(a)] / h
Tangent line at (a, f(a)):
y - f(a) = f'(a)(x - a)
Point-slope form (general):
y - y₁ = m(x - x₁)
Conjugate multiplication trick (for square roots):
Multiply numerator and denominator by √(a + h) + √a to clear the radical from the numerator.
The derivative is the mathematical version of "speedometer reading." If f(t) is your position at time t, f'(t) is your velocity at that instant. Every time an engineer, economist, or scientist talks about a rate of change (how fast a population grows, how quickly a chemical reaction proceeds, how a stock price is moving right now), they are using derivatives.
Students try to use shortcut rules (power rule, product rule) before they have been introduced. In this section, every derivative must come from the limit definition. If the problem says "use the definition," any other method earns zero credit.
When rationalising, students forget to multiply both the numerator and the denominator by the conjugate. You must do both, or you change the expression.
Students sometimes plug h = 0 directly into the difference quotient before simplifying. This gives 0/0, which is not a valid answer. The whole point is to simplify first, then take the limit.
Students confuse "not differentiable" with "not continuous." A function can be continuous at a point yet still fail to be differentiable there (corners, cusps, vertical tangents).
⚠️ "Use the definition of the derivative" is a common exam instruction. Show every step of the limit process. A bare answer with no limit work will not receive full marks.
⚠️ Tangent-line problems follow a two-step pattern: find f'(a) from the definition, then write the line in point-slope form. Practise until this is automatic.
⚠️ Expect at least one problem asking you to identify or sketch a function that is not differentiable at a specific point. Know the visual signatures: corner, cusp, vertical tangent, discontinuity.
⚠️ The relationship "differentiable ⇒ continuous" is testable. Be ready to explain it and to give a counterexample showing the converse is false (e.g., |x| at x = 0).
Fill in the blank: f'(a) = lim h→0 ____. (Answer: [f(a + h) - f(a)] / h.)
True or false: If f is differentiable at x = 5, then f is continuous at x = 5. (True.)
True or false: If f is continuous at x = 5, then f is differentiable at x = 5. (False. Counterexample: f(x) = |x - 5|.)
Fill in the blank: The tangent line at (a, f(a)) has slope ____. (Answer: f'(a).)
True or false: A function with a vertical tangent at x = 2 is differentiable at x = 2. (False. The derivative is infinite, so it does not exist as a real number.)
Q: Use the definition of the derivative to find f'(4) for f(x) = 3x + 4.
A: f'(4) = lim h→0 [(3(4+h)+4) - (3·4+4)] / h = lim h→0 [12+3h+4-16] / h = lim h→0 3h/h = 3.
Q: Use the definition of the derivative to find the equation of the tangent line to y = √x at (1, 1).
A: f'(1) = lim h→0 [√(1+h) - 1] / h. Multiply by the conjugate: = lim h→0 h / [h(√(1+h) + 1)] = lim h→0 1/(√(1+h) + 1) = 1/2. Tangent line: y - 1 = (1/2)(x - 1), so y = (1/2)x + 1/2.
Q: Use the definition of the derivative to find the tangent line to y = 3/x at (3, 1).
A: f'(3) = lim h→0 [3/(3+h) - 1] / h = lim h→0 [(3 - 3 - h)/(3+h)] / h = lim h→0 -1/(3+h) = -1/3. Tangent line: y - 1 = (-1/3)(x - 3), so y = (-1/3)x + 2.
Q: Give two examples of a function that is not differentiable at x = 4, each for a different reason.
A: (1) f(x) = |x - 4|: not differentiable at x = 4 because of a corner (the left-hand slope is -1, the right-hand slope is +1). (2) f(x) = { 2 for x < 4, 5 for x ≥ 4 }: not differentiable at x = 4 because f is discontinuous there (jump discontinuity).
The limit definition here is the foundation; the shortcut rules you learn next (power rule, product rule, quotient rule, chain rule) are all derived from it. Understanding why the definition works makes those rules feel less like arbitrary formulas. The tangent line reappears in linear approximation and differentials (Sec. 3.10), where you use the tangent line to estimate function values near a known point.
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