Introduction to Differentiation and First Principles (Polynomials) — Edexcel A-level Mathematics (Pure)

SPEC 7.1 • DIFFERENTIATION

Introduction to Differentiation and First Principles

Understanding the derivative as a limit and differentiating polynomials from first principles

The Problem of Gradients

For a straight line, the gradient is constant at every point. We can easily calculate it using:

Gradient of a Line

m = \frac{change in y}{change in x} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

But what about curves? The gradient changes at every point!

Key Question

How can we find the gradient of a curve at a specific point when the gradient is constantly changing?

The answer lies in finding the gradient of the tangent to the curve at that point. This leads us to the gradient function — a function that gives us the gradient at any point on the curve.

The Gradient as a Limit

Consider the curve

y = f (x)

and two points:

Point A at

(x, f (x))

Point B at

(x + h, f (x + h))

Gradient of Chord AB

\frac{f ( x + h ) - f ( x )}{h}

As

h \to 0

, point B approaches point A, and the chord becomes the tangent.

Definition of the Derivative

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

This process is called differentiation from first principles.

Derivative Notation

The derivative is also called the gradient function because it gives the gradient at any point on the curve.

NotationRead AsNamed After

f^{'} (x)

"f prime of x"Lagrange

\frac{d y}{d x}

"dy by dx"Leibniz

\frac{d}{d x} [f (x)]

"d by dx of f of x"Leibniz (operator)

Second Derivative

Differentiating the gradient function gives the second derivative:

f^{''} (x) or \frac{d ^{2} y}{d x ^{2}}

Read as "f double-prime" or "d two y by dx squared"

Rate of Change

\frac{d y}{d x}

represents the rate of change of

y

with respect to

x

.

First Principles Process

To differentiate

f (x)

from first principles, follow these steps:

Method

Write out

f (x + h)

by replacing every

x

with

(x + h)

Form the expression

\frac{f ( x + h ) - f ( x )}{h}

Expand and simplify the numerator

Cancel the common factor of

h

Take the limit as

h \to 0

Tip

After simplifying, you should always be able to cancel the

h

in the denominator. If you cannot, check your algebra!

Worked Example01

Differentiate $f (x) = x^{2}$ from first principles.

•

1. Find

f (x + h)

:

f (x + h) = (x + h)^{2} = x^{2} + 2 x h + h^{2}

•

2. Apply first principles formula:

•

3. Simplify the numerator:

•

4. Cancel

h

and evaluate the limit:

Answer: $f^{'} (x) = 2 x$

Worked Example02

Differentiate $f (x) = x^{3}$ from first principles.

•

1. Find

f (x + h)

:

f (x + h) = (x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}

•

2. Apply first principles formula:

•

3. Simplify and factor:

•

4. Cancel

h

and evaluate:

Answer: $f^{'} (x) = 3 x^{2}$

Worked Example03

Differentiate $f (x) = \frac{1}{x}$ from first principles.

•

1. Find

f (x + h)

:

f (x + h) = \frac{1}{x + h}

•

2. Apply first principles formula:

•

3. Combine the fractions in the numerator:

•

4. Simplify by dividing by

h

:

•

5. Evaluate the limit:

Answer: $f^{'} (x) = - \frac{1}{x ^{2}}$

Worked Example04

Differentiate $f (x) = x^{2} + 2 x$ from first principles.

•

1. Find

f (x + h)

:

f (x + h) = (x + h)^{2} + 2 (x + h) = x^{2} + 2 x h + h^{2} + 2 x + 2 h

•

2. Apply first principles formula:

•

3. Simplify and factor:

•

4. Cancel

h

and evaluate:

Answer: $f^{'} (x) = 2 x + 2$

Practice

Question 1: Differentiate

f (x) = x^{2}

from first principles and hence find

f^{'} (4)

.

From first principles (as shown in Example 01):

f (x) = x^{2} ⟹ f^{'} (x) = 2 x

Therefore:

f^{'} (4) = 2 (4) = 8

Answer: $f^{'} (x) = 2 x$ and $f^{'} (4) = 8$

Practice

Question 2: Use first principles to find the derivative of

f (x) = 5 x^{2}

.

f (x + h) = 5 (x + h)^{2} = 5 (x^{2} + 2 x h + h^{2}) = 5 x^{2} + 10 x h + 5 h^{2}

Answer: $f^{'} (x) = 10 x$

Practice

Question 3: Differentiate

f (x) = x^{3} + x

from first principles.

f (x + h) = (x + h)^{3} + (x + h) = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} + x + h

Answer: $f^{'} (x) = 3 x^{2} + 1$

Summary

Key Concepts

ConceptDescription

Derivative / Gradient FunctionThe gradient of the tangent to a curve at any point

First Principles

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

First Derivative Notation

f^{'} (x)

or

\frac{d y}{d x}

Second Derivative Notation

f^{''} (x)

or

\frac{d ^{2} y}{d x ^{2}}

Key Exam Skills

Expand

(x + h)^{n}

correctly using binomial expansion

Simplify the difference quotient by factoring out

h

Take the limit as

h \to 0

to find the gradient function

Evaluate the derivative at specific points to find gradients