Differentiation from First Principles Revision Notes for Grade 12 NSC Matric Mathematics

Differentiation from First Principles

What is differentiation from first principles?

Differentiation from first principles is the fundamental method used to find the derivative of a function using the basic definition. This approach uses limits to calculate the exact gradient of a curve at any point, rather than using pre-learned rules.

We know that the gradient of a tangent line to a curve with equation y = f(x) at point x = a can be found using the formula:

Gradient at a point = $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

This same formula can be used to create a general expression that describes the gradient at any point on the graph. This expression is called the derivative.

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The beauty of differentiation from first principles is that it provides a rigorous mathematical foundation for finding derivatives. Once you understand this method, all the shortcut rules for differentiation will make much more sense.

Key definitions

Derivative

The derivative of a function f(x) is written as f'(x) and represents the gradient of the tangent to the curve at any point x.

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

This is the most important formula in calculus - it defines what a derivative actually means mathematically.

Differentiation

Differentiation is the process of finding the derivative of a given function.

The method of using the limit definition to find derivatives is called differentiation from first principles or using the definition.

The step-by-step method

When differentiating from first principles, follow these steps:

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The Four-Step Method:

Write the formula: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
Substitute and simplify: Find f(x + h), substitute into the formula, and simplify the fraction
Take the limit: Evaluate the limit as h approaches 0
Write the final answer: State the derivative clearly

Worked examples

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Worked Example 1: Linear function

Question: Calculate the derivative of g(x) = 2x - 3 from first principles.

Solution:

Step 1: Write the formula

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

Step 2: Find g(x + h) and substitute

g(x) = 2x - 3

g(x + h) = 2(x + h) - 3 = 2x + 2h - 3

Step 3: Substitute and simplify

$g'(x) = \lim_{h \to 0} \frac{(2x + 2h - 3) - (2x - 3)}{h}$

$= \lim_{h \to 0} \frac{2h}{h}$

$= \lim_{h \to 0} 2 = 2$

Answer: g'(x) = 2

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Worked Example 2: Cubic function

Question: Find the derivative of f(x) = 4x³ from first principles and interpret f'(0.5).

Solution:

Step 1: Write the formula

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

Step 2: Substitute and simplify

$f'(x) = \lim_{h \to 0} \frac{4(x + h)³ - 4x³}{h}$

Expand $(x + h)³ = x³ + 3x²h + 3xh² + h³$ :

$= \lim_{h \to 0} \frac{4(x³ + 3x²h + 3xh² + h³) - 4x³}{h}$
$= \lim_{h \to 0} \frac{4x³ + 12x²h + 12xh² + 4h³ - 4x³}{h}$
$= \lim_{h \to 0} \frac{12x²h + 12xh² + 4h³}{h}$
$= \lim_{h \to 0} \frac{h(12x² + 12xh + 4h²)}{h}$
$= \lim_{h \to 0} (12x² + 12xh + 4h²) = 12x²$

Step 3: Calculate f'(0.5)

f'(x) = 12x²
f'(0.5) = 12(0.5)² = 12(1/4) = 3

Interpretation: The gradient of the tangent to f(x) at x = 0.5 equals 3.

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Worked Example 3: Rational function

Question: Calculate dp/dx from first principles if p(x) = -2/x.

Solution:

Step 1: Write the formula

$\frac{dp}{dx} = \lim_{h \to 0} \frac{p(x + h) - p(x)}{h}$

Step 2: Substitute and simplify

$\frac{dp}{dx} = \lim_{h \to 0} \frac{-\frac{2}{x + h} - \left(-\frac{2}{x}\right)}{h}$
$= \lim_{h \to 0} \frac{-\frac{2}{x + h} + \frac{2}{x}}{h}$

To simplify, find a common denominator:

$= \lim_{h \to 0} \frac{1}{h} \times \frac{-2x + 2(x + h)}{x(x + h)}$
$= \lim_{h \to 0} \frac{1}{h} \times \frac{-2x + 2x + 2h}{x(x + h)}$
$= \lim_{h \to 0} \frac{1}{h} \times \frac{2h}{x(x + h)}$
$= \lim_{h \to 0} \frac{2}{x(x + h)} = \frac{2}{x²}$

Answer: dp/dx = 2/x²

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Worked Example 4: Constant function

Question: Differentiate g(x) = 1/4 from first principles.

Solution:

Step 1: Write the formula

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

Step 2: Substitute and simplify

Since g(x) = 1/4 is constant, g(x + h) = 1/4:
$g'(x) = \lim_{h \to 0} \frac{\frac{1}{4} - \frac{1}{4}}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0$

Interpretation: The gradient of a constant function is always 0, meaning the graph is a horizontal line with no slope.

Different notations for derivatives

Understanding the various ways to express derivatives is crucial for mathematical literacy. There are several equivalent notations you'll encounter:

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Equivalent Derivative Notations

If y = f(x), then the derivative can be written as: $f'(x) = y' = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}[f(x)] = Df(x) = D_x y$

The symbols D and d/dx are called differential operators because they indicate the operation of differentiation.

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Important note: $\frac{dy}{dx}$ is not a fraction and does not mean dy ÷ dx. It's a single symbol representing the derivative of y with respect to x.

Key exam tips

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Critical Exam Success Tips:

Always start by writing the first principles formula - this shows the examiner you understand the method
Take your time with algebraic simplification - this is where most errors occur
Factor out h from the numerator before cancelling with the h in the denominator
Remember that the limit of a constant is just the constant
For rational functions, find common denominators carefully
Check your final answer makes sense (e.g., derivative of constant = 0)

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Key Points to Remember:

First principles formula: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
The derivative represents the gradient of the tangent at any point on the curve
Constant functions always have derivative zero
Linear functions have constant derivatives
Always simplify the fraction before taking the limit to avoid division by zero
Multiple notations exist for derivatives - learn to recognise them all

Differentiation from First Principles (Grade 12 NSC Matric Mathematics): Revision Notes