Calculus is essential in architectural design and analysis for modelling curves, slopes, areas, and changes in structural systems. Differentiation, a core concept of calculus, helps architects determine rates of change such as gradients of curves, optimize design parameters, and analyse forces on structures. In Kenya’s dynamic construction environment, mastering differentiation enables precise calculations necessary for innovative and safe architectural forms.

3.1 Differentiation of Algebraic Functions

Differentiation finds the rate at which one quantity changes with respect to another. For an algebraic function y = f(x), the derivative dy/dx gives the gradient of the curve at any point, which architects use to find slopes of roofs and ramps, rates of change of areas and volumes, and the shape of curved elements such as arches and shell roofs. A small set of rules allows any algebraic function to be differentiated without working from first principles each time.

Example: a design studio in Nanyuki differentiates the profile of a parabolic entrance canopy to find the slope of the canopy at each column, so that rainwater runs off and the gutter falls are set correctly.

3.1.1 Rules of differentiation (power, product, quotient, chain rules)

The four rules below are used together. The power rule handles single terms such as 5x⁴; the product rule handles two functions multiplied together; the quotient rule handles one function divided by another; and the chain rule handles a function inside another function, such as (3x + 2)⁴. Two further facts are used throughout: the derivative of a constant is zero, and the derivative of a sum is the sum of the derivatives.

Power rule

If y = xⁿ, where n is any real number (positive, negative or a fraction), then:

dy/dx = n × xⁿ⁻¹

If y = a × xⁿ, where a is a constant, then dy/dx = a × n × xⁿ⁻¹. A constant on its own (y = 7) has derivative 0, because a horizontal line has zero gradient.

Example 1: Find the gradient function of y = x³.

dy/dx = 3 × x³⁻¹

dy/dx = 3x²

Example 2: Differentiate y = 5x⁴.

dy/dx = 5 × 4 × x⁴⁻¹

dy/dx = 20x³

Example 3: Differentiate y = 1/x².

Rewrite as a power: y = x⁻²

dy/dx = −2 × x⁻²⁻¹ = −2x⁻³

dy/dx = −2/x³

Example 4: Differentiate y = 3x^(5/2).

dy/dx = 3 × (5/2) × x^(5/2 − 1)

dy/dx = (15/2)x^(3/2)

Example 5: Differentiate y = 7.

y is a constant, so dy/dx = 0.

Product rule

When y = u × v, where u and v are both functions of x, then:

dy/dx = u × (dv/dx) + v × (du/dx)

Find u and v, differentiate each, then substitute into the rule and simplify.

Example 1: Differentiate y = x² sin x.

u = x², so du/dx = 2x

v = sin x, so dv/dx = cos x

dy/dx = x² cos x + 2x sin x

Example 2: Differentiate y = (3x + 1)(x² − 4).

u = 3x + 1, so du/dx = 3

v = x² − 4, so dv/dx = 2x

dy/dx = (3x + 1)(2x) + (x² − 4)(3)

dy/dx = 6x² + 2x + 3x² − 12

dy/dx = 9x² + 2x − 12

Example 3: Differentiate y = x eˣ.

u = x, so du/dx = 1

v = eˣ, so dv/dx = eˣ

dy/dx = x eˣ + eˣ = eˣ(1 + x)

Example 4: Differentiate y = 2x³√x.

u = 2x³, so du/dx = 6x²

v = x^(1/2), so dv/dx = (1/2)x^(−1/2)

dy/dx = 2x³ × (1/2)x^(−1/2) + x^(1/2) × 6x²

dy/dx = x^(5/2) + 6x^(5/2)

dy/dx = 7x^(5/2)

Example 5: Differentiate y = (x² + 1)(x³ − 5x).

u = x² + 1, so du/dx = 2x

v = x³ − 5x, so dv/dx = 3x² − 5

dy/dx = (x² + 1)(3x² − 5) + (x³ − 5x)(2x)

dy/dx = 3x⁴ − 5x² + 3x² − 5 + 2x⁴ − 10x²

dy/dx = 5x⁴ − 12x² − 5

Quotient rule

When y = u/v, where u and v are functions of x, then:

dy/dx = [v × (du/dx) − u × (dv/dx)] ÷ v²

The order in the numerator matters: the bottom function times the derivative of the top, minus the top function times the derivative of the bottom.

Example 1: Differentiate y = x²/(x + 1).

u = x², so du/dx = 2x

v = x + 1, so dv/dx = 1

dy/dx = [(x + 1)(2x) − x²(1)] ÷ (x + 1)²

dy/dx = (2x² + 2x − x²) ÷ (x + 1)²

dy/dx = (x² + 2x) ÷ (x + 1)²

Example 2: Differentiate y = (3x + 2)/x².

u = 3x + 2, so du/dx = 3

v = x², so dv/dx = 2x

dy/dx = [x² × 3 − (3x + 2) × 2x] ÷ x⁴

dy/dx = (3x² − 6x² − 4x) ÷ x⁴

dy/dx = (−3x² − 4x) ÷ x⁴ = −(3x + 4)/x³

Example 3: Differentiate y = (x³ − 1)/(x² + 1).

u = x³ − 1, so du/dx = 3x²

v = x² + 1, so dv/dx = 2x

dy/dx = [(x² + 1)(3x²) − (x³ − 1)(2x)] ÷ (x² + 1)²

dy/dx = (3x⁴ + 3x² − 2x⁴ + 2x) ÷ (x² + 1)²

dy/dx = (x⁴ + 3x² + 2x) ÷ (x² + 1)²

Example 4: Differentiate y = √x/(x + 2).

u = x^(1/2), so du/dx = 1/(2√x)

v = x + 2, so dv/dx = 1

dy/dx = [(x + 2) × 1/(2√x) − √x × 1] ÷ (x + 2)²

Multiply the top and bottom by 2√x:

dy/dx = [(x + 2) − 2x] ÷ [2√x (x + 2)²]

dy/dx = (2 − x) ÷ [2√x (x + 2)²]

Example 5: Differentiate y = (2x² + 3)/(x³ − 1).

u = 2x² + 3, so du/dx = 4x

v = x³ − 1, so dv/dx = 3x²

dy/dx = [(x³ − 1)(4x) − (2x² + 3)(3x²)] ÷ (x³ − 1)²

dy/dx = (4x⁴ − 4x − 6x⁴ − 9x²) ÷ (x³ − 1)²

dy/dx = (−2x⁴ − 9x² − 4x) ÷ (x³ − 1)²

Chain rule

When y is a function of u and u is a function of x (a function of a function), then:

dy/dx = (dy/du) × (du/dx)

In practice: differentiate the outer function, leaving the inside unchanged, then multiply by the derivative of the inside.

Example 1: Differentiate y = (3x + 2)⁴.

Let u = 3x + 2, so y = u⁴

dy/du = 4u³ and du/dx = 3

dy/dx = 4(3x + 2)³ × 3 = 12(3x + 2)³

Example 2: Differentiate y = sin(2x²).

Let u = 2x², so y = sin u

dy/du = cos u and du/dx = 4x

dy/dx = 4x cos(2x²)

Example 3: Differentiate y = e^(5x³).

Let u = 5x³, so y = eᵘ

dy/du = eᵘ and du/dx = 15x²

dy/dx = 15x² e^(5x³)

Example 4: Differentiate y = √(4x² + 1).

Let u = 4x² + 1, so y = u^(1/2)

dy/du = (1/2)u^(−1/2) and du/dx = 8x

dy/dx = (1/2)(4x² + 1)^(−1/2) × 8x

dy/dx = 4x ÷ √(4x² + 1)

Example 5: Differentiate y = ln(x² + 3x + 2).

Let u = x² + 3x + 2, so y = ln u

dy/du = 1/u and du/dx = 2x + 3

dy/dx = (2x + 3) ÷ (x² + 3x + 2)

Choosing the right rule

• Single power term: use the power rule, after rewriting roots and fractions as powers (√x = x^(1/2), 1/x³ = x⁻³).
• Two functions multiplied: use the product rule, for example x² sin x or (2x + 1)(x³ − 4).
• One function divided by another: use the quotient rule, or rewrite as a product with a negative power and use the product rule.
• Bracket raised to a power, or a function of a function: use the chain rule, for example (5x − 1)⁶ or sin(3x²).
• Combined forms: apply the rules in layers, for example x²(3x + 1)⁵ needs the product rule with the chain rule inside it.

Chapter Summary

This chapter covered the application of calculus with a focus on differentiation and integration techniques essential for architects. It began with the differentiation of algebraic functions, emphasizing the power, product, quotient, and chain rules. The study then extended to differentiating trigonometric and hyperbolic functions, including sine, cosine, tangent, and their hyperbolic counterparts. Inverse trigonometric functions such as arcsin, arccos, and arctan were also differentiated to understand their rates of change. The concept of rate of change was explored further through first and second derivatives, highlighting practical uses in velocity and acceleration. Stationary points were analyzed to identify maxima, minima, and points of inflection, which are critical in solving optimization problems. Integration techniques were introduced for algebraic functions, covering both definite and indefinite integrals with applications in calculating areas. Finally, integration of trigonometric, hyperbolic, and logarithmic functions was examined, along with applications in finding areas under curves and volumes of solids of revolution.