Engineering mathematics is essential for professionals across all sectors in Kenya, from healthcare to agriculture and business. Understanding derivatives allows for analysis of rates of change, which is critical in optimizing processes, managing resources, and making informed decisions. Differentiation from first principles provides the foundational understanding of how derivatives are defined and calculated, forming the basis for more advanced mathematical applications.

6.1 Definition of derivatives of a function

The derivative of a function measures how the function's output changes as its input changes. In practical terms, it represents the rate of change or slope of the function at any given point. For example, in a county hospital, the rate at which patient numbers increase or decrease over time can be modeled by a derivative.

The derivative \( f'(x) \) of a function \( f(x) \) at a point \( x \) is defined by the limit:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

This limit, if it exists, gives the instantaneous rate of change of \( f(x) \) with respect to \( x \).

Worked Examples

Example 1: A retail business models its daily sales \( S(t) = 50t^2 \) where \( t \) is the number of days. Find the rate of change of sales on day 3.

Given: \( S(t) = 50t^2 \), \( t = 3 \)

$$ S'(t) = \lim_{h \to 0} \frac{50(t + h)^2 - 50t^2}{h} $$

$$ = \lim_{h \to 0} \frac{50(t^2 + 2th + h^2) - 50t^2}{h} $$

$$ = \lim_{h \to 0} \frac{50(2th + h^2)}{h} $$

$$ = \lim_{h \to 0} 50(2t + h) $$

Substitute \( t = 3 \):

$$ = 50(2 \times 3 + 0) = 50 \times 6 = 300 $$

Answer: The rate of change of sales on day 3 is 300 units per day.

Example 2: A university models the number of new student enrollments as \( E(x) = 1000 + 20x - 0.5x^2 \), where \( x \) is the number of months since January. Find the rate of change of enrollments at \( x = 4 \).

Given: \( E(x) = 1000 + 20x - 0.5x^2 \), \( x = 4 \)

$$ E'(x) = \lim_{h \to 0} \frac{(1000 + 20(x+h) - 0.5(x+h)^2) - (1000 + 20x - 0.5x^2)}{h} $$

$$ = \lim_{h \to 0} \frac{20h - 0.5[(x+h)^2 - x^2]}{h} $$

Expand \( (x+h)^2 - x^2 = 2xh + h^2 \):

$$ = \lim_{h \to 0} \frac{20h - 0.5(2xh + h^2)}{h} = \lim_{h \to 0} \frac{20h, xh - 0.5h^2}{h} $$

$$ = \lim_{h \to 0} (20 - x - 0.5h) = 20 - x $$

Substitute \( x = 4 \):

$$ = 20 - 4 = 16 $$

Answer: The rate of change of enrollments at month 4 is 16 students per month.

Example 3: The water level \( W(t) \) in a dam is given by \( W(t) = 5t^3 - 12t^2 + 7t \), where \( t \) is time in hours. Find the rate of change of water level at \( t = 2 \) hours.

Given: \( W(t) = 5t^3 - 12t^2 + 7t \), \( t = 2 \)

$$ W'(t) = \lim_{h \to 0} \frac{5(t+h)^3 - 12(t+h)^2 + 7(t+h) - (5t^3 - 12t^2 + 7t)}{h} $$

Expand the terms:

$$ 5[(t+h)^3 - t^3] - 12[(t+h)^2 - t^2] + 7h $$

Using binomial expansions:

$$ (t+h)^3 - t^3 = 3t^2h + 3th^2 + h^3 $$

$$ (t+h)^2 - t^2 = 2th + h^2 $$

Substitute:

$$ = \lim_{h \to 0} \frac{5(3t^2h + 3th^2 + h^3) - 12(2th + h^2) + 7h}{h} $$

$$ = \lim_{h \to 0} \frac{15t^2h + 15th^2 + 5h^3 - 24th - 12h^2 + 7h}{h} $$

Group terms:

$$ = \lim_{h \to 0} \frac{(15t^2h - 24th + 7h) + (15th^2 - 12h^2) + 5h^3}{h} $$

Divide each term by \( h \):

$$ = \lim_{h \to 0} (15t^2 - 24t + 7) + (15t - 12)h + 5h^2 $$

As \( h \to 0 \):

$$ = 15t^2 - 24t + 7 $$

Substitute \( t = 2 \):

$$ = 15 \times 2^2 - 24 \times 2 + 7 = 15 \times 4 - 48 + 7 = 60 - 48 + 7 = 19 $$

Answer: The rate of change of water level at 2 hours is 19 units per hour.

Chapter Summary

This chapter began with the definition of derivatives of a function and the foundational concept of differentiation from first principles. It introduced tables of some common derivatives to facilitate quick reference. The chapter then detailed the various rules of differentiation, including sum and difference, product, quotient, chain, parametric, and implicit differentiation techniques. The concepts of rate of change and small change were explored to highlight their practical significance. Stationary points of functions of two variables were examined to understand points of maxima, minima, and saddle points. Integration was defined and distinguished between indefinite and definite integrals. Various methods of integration such as substitution, reduction formula, integration by parts, and partial fractions were explained alongside integrals of hyperbolic and inverse functions. Finally, the chapter emphasized the application of integration in solving engineering problems.