8.1 Differentiation

Key Takeaways

  • The derivative f′(a) is the limit of [f(a+h) − f(a)]/h as h → 0—the instantaneous rate of change and the slope of the tangent at x = a.
  • Power rule: d/dx[xⁿ] = n xⁿ⁻¹; product: (uv)′ = u′v + uv′; quotient: (u/v)′ = (u′v − uv′)/v²; chain: d/dx[f(g(x))] = f′(g(x))·g′(x).
  • Tangent line at x = a has slope f′(a) and equation y − f(a) = f′(a)(x − a).
  • Related rates link dy/dt to dx/dt via the chain rule; always differentiate both sides with respect to time before substituting known values.
  • Practise derivative evaluations, tangent slopes, and one-step rate problems after learning the underlying rules.
Last updated: July 2026

Why Differentiation Matters

Differentiation is foundational mathematics for technical study. This independent review emphasises fluent derivative rules rather than formal ε–δ arguments. Current public PAF sources do not publish a mathematics weighting, question count, timing, or confirmed subject allocation for this route.

This section covers the FSc Pre-Engineering toolkit: limits as the idea behind the derivative, the power, product, quotient, and chain rules, tangent lines, and rates of change. Integration and probability follow in later sections of this chapter.

Limits: The Idea Behind the Derivative

The average rate of change of y = f(x) from x = a to x = a + h is

[\frac{f(a+h) - f(a)}{h}.]

The derivative at x = a is the limit of that quotient as h approaches 0 (when the limit exists):

[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.]

Geometrically, f′(a) is the slope of the tangent to the curve at the point (a, f(a)). Physically, if s(t) is position, s′(t) is instantaneous velocity.

Worked limit definition. Let f(x) = x². Then

[f'(3) = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0}(6 + h) = 6.]

So the tangent slope at x = 3 is 6. For efficient calculation you will usually use the rules below instead of expanding the difference quotient—but knowing the definition explains why those rules exist.

One-sided / continuity note (MCQ trap). Differentiability at a requires the two-sided limit of the difference quotient to exist, which forces continuity at a. Continuity alone does not guarantee differentiability (classic corner: f(x) = |x| at 0).

Standard Derivative Rules (Memorize Cold)

RuleFormulaRecognition cue
Constantd/dx[c] = 0Constant term vanishes
Powerd/dx[xⁿ] = n xⁿ⁻¹n any real (FSc: integer/rational)
Constant multipled/dx[c f] = c f′Factor out coefficients
Sum/difference(u ± v)′ = u′ ± v′Term-by-term
Product(uv)′ = u′v + uv′Two factors
Quotient(u/v)′ = (u′v − uv′)/v²Fraction of functions
Chaind/dx[f(g(x))] = f′(g(x))·g′(x)Composition / “outer × inner”

Common FSc forms:

  • d/dx[sin x] = cos x, d/dx[cos x] = −sin x, d/dx[tan x] = sec² x
  • d/dx[eˣ] = eˣ, d/dx[ln x] = 1/x (x > 0)
  • d/dx[aˣ] = aˣ ln a (when a > 0, a ≠ 1)

Worked power rule. Differentiate y = 5x⁴ − 3x² + 7. Then y′ = 20x³ − 6x. At x = 1, y′(1) = 14.

Worked product rule. Let u = x², v = sin x. Then y = x² sin x gives

y′ = 2x sin x + x² cos x.

At x = π/2: y′(π/2) = 2·(π/2)·1 + (π/2)²·0 = π.

Worked quotient rule. Let y = (3x + 1)/(x − 2). Then u = 3x + 1, v = x − 2,

u′ = 3, v′ = 1, so

y′ = [3(x − 2) − (3x + 1)(1)] / (x − 2)² = (3x − 6 − 3x − 1)/(x − 2)² = −7/(x − 2)².

At x = 0, y′(0) = −7/4.

Worked chain rule. Differentiate y = (2x³ − 5)⁵. Outer power 5, inner g = 2x³ − 5:

y′ = 5(2x³ − 5)⁴ · 6x² = 30x²(2x³ − 5)⁴.

At x = 1: y′(1) = 30(2 − 5)⁴ = 30 · 81 = 2430.

Nested chain example. y = sin(3x²). Then y′ = cos(3x²) · 6x = 6x cos(3x²).

Tangents

If f is differentiable at a, the tangent line at (a, f(a)) is

[y - f(a) = f'(a),(x - a).]

Worked tangent. For f(x) = x³ − 2x at a = 2: f(2) = 8 − 4 = 4, f′(x) = 3x² − 2, f′(2) = 12 − 2 = 10. Tangent:

y − 4 = 10(x − 2) ⇒ y = 10x − 16.

Normal line (occasional): slope −1/f′(a) when f′(a) ≠ 0. At the point above, normal slope = −1/10, so y − 4 = (−1/10)(x − 2).

Rates of Change and Related Rates

If a quantity Q depends on time t, dQ/dt is its instantaneous rate. When two variables are linked by an equation, differentiate both sides with respect to t (chain rule), then substitute known values.

Worked related rates. A circular oil slick has radius r(t) with area A = πr². If r increases at 0.2 m/s when r = 5 m, find dA/dt.

Differentiating: dA/dt = 2πr · dr/dt = 2π·5·0.2 = 2π m²/s ≈ 6.28 m²/s.

Ladder / sliding object pattern. A 13 m ladder leans against a wall; base moves away at 0.5 m/s when base is 5 m from the wall. Then x² + y² = 169. Differentiate: 2x dx/dt + 2y dy/dt = 0 ⇒ dy/dt = −(x/y) dx/dt. When x = 5, y = 12, so dy/dt = −(5/12)(0.5) = −5/24 m/s (negative = descending).

Practice Focus

Expect stems such as “f′(2) = ?”, “slope of the tangent at …”, “differentiate (3x − 1)⁴”, and short related-rate numbers. Skip multi-step geometry rates if a pure rule question is clearer—return only if time remains. Always simplify before plugging x = a when the algebra is messy; many wrong options are “forgot the chain factor” or “used product instead of quotient.”

Test Your Knowledge

Using the definition or power rule, what is f′(3) if f(x) = x²?

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Test Your Knowledge

If y = (3x + 1)/(x − 2), what is y′ as a simplified function of x?

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Test Your Knowledge

For f(x) = x³ − 2x, what is the equation of the tangent line at x = 2?

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Test Your Knowledge

A circular region has area A = πr². If dr/dt = 0.2 m/s when r = 5 m, what is dA/dt?

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