Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

Lessons

Senior 4 · Mathematics

Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

5 lessons

Lesson 1 of 5

The derivative: definition, first principles and the power rule

25 min

From average to instantaneous

In Unit 7 a falling stone travelled $s (t) = 5 t^{2}$ metres in $t$ seconds, and its speed at $t = 2$ came out as the limit of average speeds, $20\,\mathrm{m\,s^{-1}}$ . The same limit gives the steepness of a curve at a point: the slope of a chord through two points of the curve tends to the slope of the tangent as the points come together. This single limit, the derivative, measures how fast any quantity changes: speed, the growth of a population, the marginal cost of producing one more item.

Chords from $P (1, 1)$ to $Q (1 + h, (1 + h)^{2})$ for $h = 1.5, 1, 0.5$ have slopes $3.5, 3, 2.5$ . As $h \to 0$ they turn into the tangent at $P$ , of slope 2 (red).

Definition

Derivative of $f$ at $a$

f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

provided this limit exists (as a finite number); $f$ is then differentiable at $a$ . The fraction is the slope of the chord from $(a, f (a))$ to $(a + h, f (a + h))$ ; its limit $f^{'} (a)$ is the slope of the tangent at $(a, f (a))$ , and the instantaneous rate of change of $f$ at $a$ .

Derivative function: $f^{'}$ , which gives $f^{'} (x)$ at each $x$ where $f$ is differentiable. Notations: $f^{'} (x)$ , $y^{'}$ , $\frac{dy}{dx}$ , $\frac{d}{dx}f(x)$ . Finding it is called differentiation; using the definition directly is differentiating from first principles.

Worked example 1: $x^{2}$ and $x^{3}$ from first principles

Differentiate (a) $f (x) = x^{2}$ (b) $g (x) = x^{3}$ .

Solution

$\frac{(x + h)^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h \to 2x$ . So $f^{'} (x) = 2 x$ ; at $x = 1$ the slope is 2, as in the diagram.
$\frac{(x + h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \to 3x^2$ . So $g^{'} (x) = 3 x^{2}$ .

Worked example 2: a rational and an irrational function

Differentiate from first principles (a) $f(x) = \frac{1}{x}$ , $x \ne 0$ (b) $g(x) = \sqrt{x}$ , $x \gt 0$ .

Solution

$\frac{1}{h}\left(\frac{1}{x + h} - \frac{1}{x}\right) = \frac{1}{h} \cdot \frac{x - (x + h)}{x(x + h)} = \frac{-1}{x(x + h)} \to -\frac{1}{x^2}$ .
Multiply by the conjugate (Unit 7): $\frac{\sqrt{x + h} - \sqrt{x}}{h} = \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{1}{\sqrt{x + h} + \sqrt{x}} \to \frac{1}{2\sqrt{x}}$ .

Important

Differentiable implies continuous. If $f^{'} (a)$ exists, then $f(a + h) - f(a) = \frac{f(a + h) - f(a)}{h} \cdot h \to f'(a) \times 0 = 0$ , so $f$ is continuous at $a$ .

The converse is false. $f (x) = ∣ x ∣$ is continuous at 0, but $\frac{|0 + h| - 0}{h} = \frac{|h|}{h}$ tends to $- 1$ from the left and to $1$ from the right (Unit 7): the left and right derivatives differ, so $∣ x ∣$ has no derivative at 0. Its graph has a corner there. And $\sqrt{x}$ at 0 has $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}} \to +\infty$ : a vertical tangent, so no derivative either.

Important

Basic rules ( $c$ and $k$ are constants, $u$ and $v$ differentiable functions):

Function	Derivative
$c$	$0$
$x^{n}$ ( $n$ rational)	$nx^{n-1}$ (power rule)
$k u$	$k u^{'}$
$u \pm v$	$u' \pm v'$ (sum and difference rules)

For a positive whole number $n$ , the binomial theorem gives $(x + h)^n = x^n + nx^{n-1}h + (\text{terms with } h^2, h^3, \ldots)$ , so the difference quotient tends to $nx^{n-1}$ . Worked example 2 checks the rule for $n = - 1$ and $n = \frac{1}{2}$ ; it holds for every rational $n$ .

Worked example 3: using the rules

Differentiate (a) $y = 4 x^{3} - 5 x^{2} + 7 x - 2$ (b) $y = 3\sqrt{x} + \frac{2}{x^2}$ (c) $y = \frac{x^3 - 2x}{x}$ , $x \ne 0$ (d) $y = (x^{2} + 1)^{2}$ .

Solution

$y^{'} = 12 x^{2} - 10 x + 7$ .
Write $y = 3x^{\frac{1}{2}} + 2x^{-2}$ : $y' = \frac{3}{2}x^{-\frac{1}{2}} - 4x^{-3} = \frac{3}{2\sqrt{x}} - \frac{4}{x^3}$ .
Simplify first: $y = x^{2} - 2$ , so $y^{'} = 2 x$ (for $x \ne 0$ ).
Expand first: $y = x^{4} + 2 x^{2} + 1$ , so $y^{'} = 4 x^{3} + 4 x$ .

Warning

The derivative of a product is not the product of the derivatives: for $y = x^2 \cdot x^3 = x^5$ , $y^{'} = 5 x^{4}$ , not $2x \cdot 3x^2 = 6x^3$ . Until Lesson 2, expand or simplify first. And rewrite roots and fractions as powers before differentiating: $\frac{2}{x^2} = 2x^{-2}$ , not $\frac{2}{2x}$ .

Tip

Read $\frac{dy}{dx}$ as "the rate of change of $y$ with respect to $x$ ". Its units are the units of $y$ per unit of $x$ : metres per second, francs per item, degrees per minute. Checking units catches many errors in applications.

3Blue1Brown: The paradox of the derivative (about 17 min). Watch how "speed at an instant" is made precise as the limit of the change in distance divided by a tiny change in time, and how that leads to the derivative of t³.

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Practice

Questions 1 to 3 must be done from first principles.

Differentiate $f (x) = 3 x^{2} - x$ from first principles.
Differentiate $f(x) = \frac{1}{x + 2}$ from first principles.
Differentiate $f(x) = \sqrt{2x + 1}$ from first principles.
Differentiate $y = 2x^5 - \frac{1}{3}x^3 + 4x - 9$ .
Differentiate $y = \frac{6}{x^3} - 4\sqrt[3]{x}$ .
Differentiate $y = \frac{x^2 + 3x - 4}{\sqrt{x}}$ by first writing it as a sum of powers.
Find the slope of the tangent to $y = x^{3} - 3 x$ at the point where $x = 2$ , and find the points where the tangent is horizontal.
Show that $f (x) = ∣ x - 3 ∣$ is continuous but not differentiable at 3.

Show answersHide answers

$\frac{3(x + h)^2 - (x + h) - 3x^2 + x}{h} = 6x + 3h - 1 \to 6x - 1$ .
$\frac{1}{h}\left(\frac{1}{x + h + 2} - \frac{1}{x + 2}\right) = \frac{-1}{(x + h + 2)(x + 2)} \to -\frac{1}{(x + 2)^2}$ .
Conjugate: $\frac{2h}{h\left(\sqrt{2x + 2h + 1} + \sqrt{2x + 1}\right)} \to \frac{1}{\sqrt{2x + 1}}$ .
$y^{'} = 10 x^{4} - x^{2} + 4$ .
$y = 6x^{-3} - 4x^{\frac{1}{3}}$ : $y' = -\frac{18}{x^4} - \frac{4}{3}x^{-\frac{2}{3}}$ .
$y = x^{\frac{3}{2}} + 3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}$ : $y' = \frac{3}{2}x^{\frac{1}{2}} + \frac{3}{2}x^{-\frac{1}{2}} + 2x^{-\frac{3}{2}}$ .
$y^{'} = 3 x^{2} - 3$ ; at $x = 2$ the slope is 9. Horizontal where $3 x^{2} - 3 = 0$ : $(1, - 2)$ and $(- 1, 2)$ .
$\lim_{x \to 3} |x - 3| = 0 = f(3)$ , so continuous. The difference quotient $\frac{|h|}{h}$ has left limit $- 1$ and right limit $1$ , so no derivative.

Quick check

Which limit defines f′(a)?

Key points

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ : the slope of the tangent and the instantaneous rate of change.
From first principles: expand or use a conjugate or a common denominator, cancel $h$ , then let $h \to 0$ .
Differentiable $\Rightarrow$ continuous, not conversely ( $∣ x ∣$ at 0).
$(x^n)' = nx^{n-1}$ for rational $n$ ; constants vanish; derivatives of sums are sums of derivatives.

Sources and credits

Written by Ganzaa: the explanations, the graph, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Image: Portrait of Sir Isaac Newton, 1689, Godfrey Kneller, Wikimedia Commons, public domain.
Video: The paradox of the derivative | Chapter 2, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1From average to instantaneous

From average to instantaneous