Apply calculus in professional contexts such as data analysis, optimization, and modeling is essential across various Kenyan industries including healthcare, agriculture, finance, and administration. Calculus provides tools to analyze changing quantities, optimize resources, and predict trends that are vital for efficient decision-making. This chapter focuses on differentiation up to the third order, covering fundamental concepts, rules, and applications relevant to diverse professional fields.

4.1 Differentiation up to third order

Differentiation measures how a quantity changes when the quantity it depends on changes. Engineers use it to find speed from distance, current from charge, the rate at which a cost or a temperature rises, and the settings at which a machine gives maximum or minimum output. This topic builds from the common types of function, through the rules that make differentiation efficient, to stationary points and rates of change, including second and third derivatives.

Notation and higher-order derivatives

The derivative of f(x) with respect to x is written f'(x) or df/dx, and it gives the instantaneous rate of change of f at x. The second derivative f''(x) is the derivative of f'(x), and the third derivative f'''(x) is the derivative of f''(x). In motion problems, if s is distance, then s' is velocity, s'' is acceleration and s''' is the rate of change of acceleration.

The first derivative is defined by a limit.

Formula: f'(x) = lim (h → 0) [f(x + h) − f(x)] ÷ h

4.1.1 Functions

Differentiation applies to functions, which describe how one variable depends on another. A function f(x) gives exactly one output for each input x, and the form of the function decides which derivative formula is used. The standard derivatives below are used throughout this topic. Example: a workshop models the cost of hiring a machine, the voltage in a circuit and the output of a sensor, and each model needs a different type of function.

4.1.1.1 Linear

A linear function has the form f(x) = ax + b, and its graph is a straight line. Its derivative is the constant a, which is the gradient, so the rate of change never varies. Example: a cable contractor buys cable at a fixed charge of Ksh 500 plus Ksh 120 per metre, so C(x) = 120x + 500 and the cost rises by Ksh 120 for every extra metre.

• Constant term: The derivative of a constant is zero.
• Gradient: The derivative of ax is a.
• Higher derivatives: The second and third derivatives of a linear function are zero.
• Meaning: The rate of change is the same at every point on the line.
• Graph: The tangent to a straight line is the line itself.

Worked example: find the rate of change of the cost C(x) = 120x + 500.

Formula: C'(x) = a

Substitution: C'(x) = 120

Working: the constant 500 differentiates to 0, and 120x differentiates to 120.

Result: the cost rises by Ksh 120 per metre at every length.

4.1.1.2 Trigonometric

Trigonometric functions such as sin x and cos x describe repeating patterns, such as alternating voltage, vibration and rotation. With x in radians, d/dx (sin x) = cos x and d/dx (cos x) = −sin x. Example: an electrical technician models a signal as v = 10 sin t volts, and its rate of change is 10 cos t volts per second.

• Sine: The derivative of sin x is cos x.
• Cosine: The derivative of cos x is −sin x.
• Radians: The formulas hold only when the angle is in radians.
• Constants: A constant multiplier stays in front of the derivative.
• Higher derivatives: The derivatives repeat in a cycle of four: sin x, cos x, −sin x, −cos x.

Worked example: differentiate f(x) = 3 sin x + 2 cos x and find f'(0).

Formula: f'(x) = 3 cos x − 2 sin x

Substitution: f'(0) = 3 cos 0 − 2 sin 0

Working: 3 × 1 − 2 × 0 = 3

Result: f'(0) = 3

4.1.1.3 Logarithmic

A logarithmic function f(x) = ln x rises quickly at first and then slowly, which suits a sensor whose response flattens at high input. Its derivative is 1 ÷ x. Example: a sensor reading is y = ln(x + 1), where x is the input level, and the rate of change of the reading falls as the input increases.

• Natural logarithm: The derivative of ln x is 1 ÷ x, for x greater than 0.
• Shifted argument: The derivative of ln(x + 1) is 1 ÷ (x + 1).
• Decreasing rate: The derivative becomes smaller as x increases, so the curve flattens.
• Domain: The function is defined only for positive arguments.
• Second derivative: The second derivative of ln x is −1 ÷ x², which is negative, so the curve is concave down.

Worked example: find y' for y = ln(x + 1), and evaluate it at x = 9.

Formula: y' = 1 ÷ (x + 1)

Substitution: y'(9) = 1 ÷ (9 + 1)

Working: 1 ÷ 10 = 0.1

Result: y'(9) = 0.1 reading units per input unit

4.1.1.4 Exponential

An exponential function f(x) = eᵏˣ models growth and decay, and its derivative is k × eᵏˣ, so the rate of change is proportional to the quantity itself. Example: the voltage across a discharging capacitor falls as V(t) = 100e^(−0.2t) volts, where t is in seconds.

• Derivative rule: The derivative of eᵏˣ is k eᵏˣ.
• Growth: A positive k gives a rate of change that is positive and increasing.
• Decay: A negative k gives a rate of change that is negative and shrinking in size.
• Proportionality: The derivative is always k times the original function.
• Higher derivatives: The nth derivative of eᵏˣ is kⁿ eᵏˣ.

Worked example: find V'(t) and evaluate it at t = 5.

Formula: V'(t) = 100 × (−0.2) × e^(−0.2t) = −20e^(−0.2t)

Substitution: V'(5) = −20e^(−1)

Working: e^(−1) = 0.36788, so −20 × 0.36788 = −7.358

Result: the voltage is falling at about 7.36 V per second at t = 5 s.

4.1.2 Rules

Differentiation rules allow the derivative of a complex function to be built from the derivatives of simple ones. The four rules in this section cover powers, products, composite functions and quotients, and most functions met in engineering can be handled with them. Example: the power in a circuit, the deflection of a beam and the average cost of a batch all need one or more of these rules.

4.1.2.1 Power

For f(x) = xⁿ, where n is any real number, d/dx (xⁿ) = n × x^(n−1). A constant multiplier stays in front, and terms in a sum are differentiated one at a time. Example: the output of a machine grows as 7x⁵.

Formula: f'(x) = 7 × 5 × x⁴

Result: f'(x) = 35x⁴

Worked example 1: a vehicle moves a distance s(t) = 5t² + 3t + 7 metres in t seconds. Find its velocity at t = 4 s.

Formula: s'(t) = 10t + 3

Substitution: s'(4) = 10(4) + 3

Working: 40 + 3 = 43

Result: velocity = 43 m/s

Worked example 2: a factory models its profit as R(x) = 2x³ − 5x² + 4x − 1 thousand Ksh, where x is the number of units sold. Find R'(x).

Formula: R'(x) = 6x² − 10x + 4

Result: R'(x) = 6x² − 10x + 4

Worked example 3: the sag of a cable is d(t) = t^(1/2) millimetres at time t minutes. Find d'(9).

Formula: d'(t) = ½ × t^(−1/2)

Substitution: d'(9) = ½ × 1 ÷ √9 = ½ × 1/3

Result: d'(9) = 1/6 mm per minute

4.1.2.2 Product

For f(x) = u(x) × v(x), f'(x) = u'(x) × v(x) + u(x) × v'(x). The rule is used when two changing quantities are multiplied, such as voltage and current in a power calculation. Example: the power in a circuit is P = u × v, with u = 3x² and v = 2x + 1.

Substitution: u' = 6x and v' = 2, so f'(x) = 6x(2x + 1) + 3x² × 2

Working: 12x² + 6x + 6x² = 18x² + 6x

Result: f'(x) = 18x² + 6x

4.1.2.3 Chain

For a composite function f(x) = g(h(x)), f'(x) = g'(h(x)) × h'(x). The rule is used when one function sits inside another, so the outer function is differentiated first and then multiplied by the derivative of the inner function. Example: a technician models a vibration signal as sin(3x² + 2).

Substitution: g' = cos(3x² + 2) and h' = 6x

Result: f'(x) = 6x cos(3x² + 2)

A second case is f(x) = e^(2x³), which gives f'(x) = e^(2x³) × 6x² = 6x² e^(2x³).

4.1.2.4 Quotient

For f(x) = u(x) ÷ v(x), f'(x) = [u'(x) × v(x) − u(x) × v'(x)] ÷ [v(x)]². The rule is used when one changing quantity is divided by another, such as a cost per unit. Example: a workshop studies an average cost with u = 5x³ + 2 and v = x².

Substitution: u' = 15x² and v' = 2x, so f'(x) = [15x² × x² − (5x³ + 2) × 2x] ÷ x⁴

Working: (15x⁴ − 10x⁴ − 4x) ÷ x⁴ = (5x⁴ − 4x) ÷ x⁴ = 5 − 4 ÷ x³

Result: f'(x) = 5 − 4 ÷ x³

4.1.3 Applications

Differentiation is used to find rates of change, to locate the best operating point of a system and to study how a quantity is changing. Higher-order derivatives give acceleration and concavity, which are useful in motion, control and forecasting. Example: an engineer finds the load at which a generator gives its maximum power, and then checks how quickly the power changes near that load.

4.1.3.1 Stationary points

Stationary points are points where f'(x) = 0, so the tangent is horizontal. They are maxima, minima or points of inflection, and they are used to optimise output, cost and efficiency. The second derivative tells the type: if f''(x) > 0 the point is a minimum, and if f''(x) < 0 it is a maximum. If f''(x) = 0, the third derivative helps: a non-zero f'''(x) shows that the concavity changes, which is a point of inflection. Example: the power delivered by a generator is P(x) = −2x² + 8x + 10 watts, where x is the load current in amperes.

Worked example 1: find the current that maximises power, and the maximum power.

Formula: P'(x) = −4x + 8, set to zero

Substitution: −4x + 8 = 0

Working: 4x = 8, so x = 2, and P''(x) = −4, which is less than 0

Result: maximum at x = 2 A, where P(2) = −8 + 16 + 10 = 18 W

Worked example 2: a factory has a cost function C(x) = 100x − 4x² + 0.05x³ thousand Ksh, where x is the number of units. Find where the concavity changes.

Formula: C''(x) = −8 + 0.3x, set to zero

Working: x = 8 ÷ 0.3 = 26.67, and C'''(x) = 0.3, which is not zero

Result: a point of inflection at about 26.7 units

4.1.3.2 Rates of change

The first derivative gives the rate at which a quantity changes per unit of time or other variable. The second derivative gives the acceleration of the quantity, and the third derivative gives the rate of change of that acceleration. Example: using the generator power above, the rate of change of power with load current can be found at any load.

Worked example 1: first derivative. Find the rate of change of P(x) = −2x² + 8x + 10 at x = 3 A.

Formula: P'(x) = −4x + 8

Substitution: P'(3) = −4(3) + 8

Working: −12 + 8 = −4

Result: power falls by 4 W per ampere at a load of 3 A

Worked example 2: up to the third derivative. The cost of running a plant is C(t) = 50 + 10t − 0.5t² + 0.02t³ thousand Ksh, where t is in months. Find the acceleration of cost at t = 5.

Formula: C'(t) = 10 − t + 0.06t², C''(t) = −1 + 0.12t, C'''(t) = 0.12

Substitution: C''(5) = −1 + 0.12 × 5

Working: −1 + 0.6 = −0.4

Result: C''(5) = −0.4 thousand Ksh per month², and C'''(5) = 0.12 thousand Ksh per month³

Worked example 3: slowest rise. The water level in a tank is h(t) = 1000 + 50t − 2t² + 0.1t³ millimetres, where t is in minutes. Find when the rate of rise is stationary, and its value.

Formula: h'(t) = 50 − 4t + 0.3t², and h''(t) = −4 + 0.6t, set to zero

Working: t = 4 ÷ 0.6 = 6.67, and h'''(t) = 0.6, which is positive, so the rate is a minimum. Then h'(6.67) = 50 − 26.67 + 13.33 = 36.67

Result: the level rises most slowly at about 6.67 minutes, at about 36.7 mm per minute

Chapter Summary

This chapter introduced differentiation up to the third order, beginning with various functions including linear, trigonometric, logarithmic, and exponential types. It covered essential differentiation rules such as the power, product, chain, and quotient rules, enabling the calculation of derivatives for complex expressions. Applications of differentiation were explored through stationary points and rates of change, highlighting how calculus can analyze variable behavior. The integration section started with standard and definite integrals, then progressed to advanced techniques including integration by parts, substitution, and partial fractions. Practical applications of integration were demonstrated by finding areas under and between curves as well as volumes of solids. Further topics included solving differential equations, evaluating double and triple integrals, and introducing integral transforms such as the Laplace transform and Fourier series, which are crucial for engineering analysis. This comprehensive treatment equips students with the calculus tools necessary for solving a wide range of engineering problems.