Limits and First Principles | IEB Grade 12 Maths

First principles is where calculus starts: you find the gradient of a curve at a single point by shrinking the gap between two points to nothing. It appears in Paper 1 almost every year, usually as a 4 to 6 mark question, and the notation is marked as carefully as the algebra.

Average gradient

The average gradient of a curve between $x = a$ and $x = b$ is the gradient of the straight line (the secant) joining the two points:

$average gradient = \frac{f ( b ) - f ( a )}{b - a}$

It tells you the average rate of change over that interval.

Example. For $f (x) = x^{2} + 1$ between $x = 1$ and $x = 3$ :

$\frac{f ( 3 ) - f ( 1 )}{3 - 1} = \frac{10 - 2}{2} = 4$

If the two points are $x$ and $x + h$ , the average gradient is $\frac{f ( x + h ) - f ( x )}{h}$ . This expression is the starting point for first principles.

Intuitive limits

A limit describes the value an expression approaches as the variable gets closer to some number, even if the expression cannot be evaluated at that number.

$lim_{h \to 0} \frac{h ^{2} + 3 h}{h}$

You cannot substitute $h = 0$ straight away because you would divide by zero. Factorise first, then cancel, because $h \neq = 0$ while it approaches 0:

$lim_{h \to 0} \frac{h ( h + 3 )}{h} = lim_{h \to 0} (h + 3) = 3$

Keep writing $lim_{h \to 0}$ on every line until you actually let $h$ become 0. Dropping it early costs a mark.

Derivative from first principles

The derivative is the limit of the average gradient as the gap $h$ shrinks to 0:

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

The method is always the same:

Write down the definition.
Find $f (x + h)$ by substituting $x + h$ for every $x$ .
Simplify $f (x + h) - f (x)$ . Every term without $h$ should cancel.
Factorise $h$ out of the numerator and cancel it.
Let $h \to 0$ .

Example. $f (x) = x^{2} - 3 x$ .

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

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

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

If a question says "from first principles", the rules of differentiation earn no marks.

Key terms

Secant: A straight line joining two points on a curve. Its gradient is the average gradient.
Tangent: A line that touches a curve at one point and has the same gradient as the curve there.
Limit: The value an expression approaches as the variable approaches a given number.
Derivative: The gradient of the tangent to a curve at any point, written $f^{'} (x)$ , $\frac{d y}{d x}$ or $D_{x}$ .
First principles: Finding the derivative from the limit definition rather than from the rules.

Common mistakes

Dropping $lim_{h \to 0}$ before $h$ is replaced by 0.
Writing $f (x + h) = x^{2} + h$ instead of $(x + h)^{2}$ . Substitute $x + h$ for every $x$ and use brackets.
Forgetting to subtract every term of $f (x)$ , especially constants and negative terms.
Cancelling $h$ before factorising it out of every term in the numerator.
Using the power rule when the question asks for first principles.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

knowledge

Determine the average gradient of $f (x) = x^{2} - 3 x$ between $x = 1$ and $x = 4$ .

Worked solution

Hidden - try the question first

Solution

$f (4) = 16 - 12 = 4$
$f (1) = 1 - 3 = - 2$
Average gradient $= \frac{f ( 4 ) - f ( 1 )}{4 - 1} = \frac{4 - ( - 2 )}{3}$
$= 2$

Question 2

5 marks total

routine

Determine $f^{'} (x)$ from first principles if $f (x) = 2 x^{2} - 5$ .

Worked solution

Hidden - try the question first

Solution

$f (4) = 16 - 12 = 4$
$f (x + h) = 2 (x + h)^{2} - 5 = 2 x^{2} + 4 x h + 2 h^{2} - 5$
$f (x + h) - f (x) = 4 x h + 2 h^{2}$
$f^{'} (x) = h \to 0 lim \frac{h ( 4 x + 2 h )}{h} = h \to 0 lim (4 x + 2 h)$
$f^{'} (x) = 4 x$

Question 3

6 marks total

complex

Given $g (x) = \frac{3}{x}$ , determine $g^{'} (x)$ from first principles.

Worked solution

Hidden - try the question first

Solution

$f (4) = 16 - 12 = 4$
$f (x + h) = 2 (x + h)^{2} - 5 = 2 x^{2} + 4 x h + 2 h^{2} - 5$
$= \frac{- 3 h}{x ( x + h )}$
$g^{'} (x) = h \to 0 lim \frac{- 3 h}{x ( x + h )} \times \frac{1}{h} = h \to 0 lim \frac{- 3}{x ( x + h )}$
$g^{'} (x) = - \frac{3}{x ^{2}}$

Question 4

9 marks total

problem-solving

A drone takes off from a rooftop. Its height above the rooftop after $t$ seconds is $h (t) = 40 t - 5 t^{2}$ metres, for $0 \leq t \leq 8$ .

(a) Calculate the average rate of change of the height between $t = 1$ and $t = 3$ . (3)

(b) Use first principles to determine $h^{'} (t)$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) $h (3) = 120 - 45 = 75$ and $h (1) = 40 - 5 = 35$
Average rate $= \frac{75 - 35}{3 - 1}$
$= 20$ m/s
(b) $h (t + k) = 40 t + 40 k - 5 t^{2} - 10 t k - 5 k^{2}$ , using $k$ for the small change in $t$
$h (t + k) - h (t) = 40 k - 10 t k - 5 k^{2}$
$h^{'} (t) = k \to 0 lim \frac{k ( 40 - 10 t - 5 k )}{k} = k \to 0 lim (40 - 10 t - 5 k)$
$h^{'} (t) = 40 - 10 t$
(c) $40 - 10 t = 20$
$t = 2$ seconds

FAQs

How many marks is first principles worth in IEB Maths Paper 1?

Usually 4 to 6 marks. It is one of the most predictable questions on the paper, so it is worth practising until the method is automatic.

Can I use the power rule to check my first principles answer?

Yes, and you should. Just do not present the rule as your answer. Marks are only given for the limit method when the question says first principles.

Why do I lose marks for notation in first principles?

Markers expect the limit to be written on every line until $h$ is replaced by 0, and they expect correct brackets around $x + h$ . These are easy marks to keep.

What is the difference between average gradient and the derivative?

Average gradient is the gradient between two points. The derivative is the gradient at one point, found by letting the gap between the two points shrink to zero.

Limits and differentiation from first principles

Average gradient

Intuitive limits

Derivative from first principles

Key terms

Common mistakes

Practice questions

FAQs

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