Chapter 2 — Part 1

MATHEMATICS HIP-HOP • UACE PURE MATHEMATICS

CHAPTER 02 — DIFFERENTIATION

PART 1: FIRST PRINCIPLES AND BASIC RULES

Concepts • Explanations • Worked Examples • Practice • Answers

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Learning objectives

By the end of this lesson, you should be able to explain differentiation as the gradient of a tangent and as an instantaneous rate of change; define a derivative from first principles; and apply the power, constant, constant-multiple, sum and difference rules to polynomial, fractional-power and negative-power functions.

1. What is differentiation?

For a straight line, the gradient is constant. For a curve, its steepness usually changes from point to point. Differentiation is the process used to find the gradient of a curve at a particular point. It also measures an instantaneous rate of change, such as the rate at which displacement changes with time.

If y=f(x), the derivative may be written as f′(x), y′ or dy/dx. At x=a, the gradient of the curve is f′(a). A positive derivative indicates that the function is increasing locally; a negative derivative indicates that it is decreasing locally. A zero derivative means the tangent is horizontal, but does not by itself prove that the point is a maximum or minimum.

2. The derivative from first principles

The derivative is defined by the limiting gradient of a secant line as two points on a curve approach one another:

First-principles definition
f′(x) = lim (h→0) [f(x+h) − f(x)] / h

Here, f(x+h) is the function value after x increases by h. The numerator is the change in the function, and dividing by h gives the gradient of a secant line. Taking the limit as h approaches zero gives the tangent gradient, provided the limit exists.

Important: Never substitute h=0 into the original fraction. Expand, simplify and cancel h first. Then evaluate the limit.

Worked Example 1: Differentiate f(x)=x² from first principles.

f(x+h) = (x+h)² = x² + 2xh + h².

f′(x) = lim (h→0) [(x²+2xh+h²)−x²]/h

= lim (h→0) [2xh+h²]/h

= lim (h→0) (2x+h)

Therefore, f′(x)=2x.

Worked Example 2: Differentiate f(x)=3x+5 from first principles.

f(x+h)=3(x+h)+5=3x+3h+5.

f′(x)=lim (h→0) [(3x+3h+5)−(3x+5)]/h

=lim (h→0) 3h/h = lim (h→0) 3.

Therefore, f′(x)=3.

3. Basic rules of differentiation

Rule 1: The power rule

d/dx (xⁿ) = nxⁿ⁻¹

Multiply by the power, then reduce the power by one.

Worked Example 3: Differentiate y=7x⁵.

dy/dx = 7(5x⁴) = 35x⁴.

Worked Example 4: Differentiate y=4x³−6x²+5x.

dy/dx = 12x²−12x+5.

Rule 2: The derivative of a constant

d/dx(c) = 0

A constant does not change as x changes, so its derivative is zero. For example, if y=9, then dy/dx=0.

Rule 3: The constant-multiple rule

d/dx [k f(x)] = k f′(x)

A constant multiplier remains in front while the function is differentiated. For example, if y=8x⁴, then dy/dx=8(4x³)=32x³.

Rule 4: Sum and difference rules

d/dx [f(x) ± g(x)] = f′(x) ± g′(x)

Differentiate each term separately, keeping its sign.

Worked Example 5: Differentiate y=5x⁴−3x³+2x−11.

dy/dx = 20x³−9x²+2−0 = 20x³−9x²+2.

4. Fractional and negative powers

Rewrite roots and fractions as powers before applying the power rule:

• √x = x^1/2
• 1/x = x⁻¹
• 1/x³ = x⁻³
• ∛x = x^1/3

Worked Example 6: Differentiate y=6√x.

y=6x^1/2, so dy/dx=6(1/2)x^−1/2 = 3/√x, for x>0.

Worked Example 7: Differentiate y=5/x².

y=5x⁻², so dy/dx=5(−2)x⁻³ = −10/x³, for x≠0.

Worked Example 8: Differentiate y=3x²+4/√x.

Rewrite y=3x²+4x^−1/2.

dy/dx=6x−2x^−3/2 = 6x−2/x^3/2, for x>0.

Common mistakes to avoid

1. Forgetting to reduce the power by one.
2. Differentiating a constant as though it were a variable term.
3. Losing the negative sign when differentiating a negative power.
4. Leaving a root or denominator unchanged instead of rewriting it as a power.
5. Substituting h=0 before simplifying the first-principles quotient.

5. Practice exercises

Attempt each question before revealing the answers. Show your working and simplify each derivative.

1. Using first principles, find the derivative of f(x)=x²+2x.
2. Using first principles, find the derivative of f(x)=4x−7.
3. Differentiate y=9x⁶.
4. Differentiate y=4x⁵−3x³+7x−8.
5. Differentiate y=12−5x⁴.
6. Differentiate y=3√x.
7. Differentiate y=7/x³.
8. Differentiate y=2x^3/2−5x^1/2.
9. Differentiate y=4/x²+6x³−9.
10. Find the gradient of y=x³−2x²+4x at x=2.

Reveal answers to the practice exercises

1. f′(x)=2x+2.
2. f′(x)=4.
3. dy/dx=54x⁵.
4. dy/dx=20x⁴−9x²+7.
5. dy/dx=−20x³.
6. dy/dx=3/(2√x), x>0.
7. dy/dx=−21/x⁴, x≠0.
8. dy/dx=3√x−5/(2√x), x>0.
9. dy/dx=−8/x³+18x², x≠0.
10. dy/dx=3x²−4x+4. At x=2, gradient=12−8+4=8.

Lesson summary

Differentiation gives the gradient of a curve and its instantaneous rate of change. First principles defines the derivative using a limit. The basic rules make differentiation efficient: use the power rule, differentiate constants to zero, retain constant multipliers, and differentiate sums or differences term by term. Rewrite roots and fractions as powers before applying the power rule.

END OF CHAPTER 02 — PART 1
Next: Part 2 — The Chain Rule