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 2 of 5

The product, quotient and chain rules, and higher derivatives

25 min

Rules that do the limits for you

Differentiating every function from first principles would be slow. In October 1684 Gottfried Wilhelm Leibniz published a short paper with a handful of rules (for sums, products, quotients and powers) that let anyone differentiate any algebraic expression mechanically. His notation $d x$ , $d y$ is still the one we use. This lesson proves and applies those rules, adds the chain rule for composite functions, and differentiates more than once.

Important

Product rule. If $u$ and $v$ are differentiable, then $(u v)^{'} = u^{'} v + u v^{'}$ .

Proof. Add and subtract $u (x + h) v (x)$ :

\frac{u(x + h)v(x + h) - u(x)v(x)}{h} = u(x + h)\,\frac{v(x + h) - v(x)}{h} + v(x)\,\frac{u(x + h) - u(x)}{h} .

As $h \to 0$ , $u(x + h) \to u(x)$ (a differentiable function is continuous), and the two quotients tend to $v^{'} (x)$ and $u^{'} (x)$ . ∎

Growth of the area $u v$ : $\Delta(uv) = v\,\Delta u + u\,\Delta v + \Delta u\,\Delta v$ . Divide by $\Delta x$ and let $\Delta x \to 0$ : the tiny corner disappears and $(u v)^{'} = u^{'} v + u v^{'}$ .

Worked example 1: products

Differentiate (a) $y = (2 x + 1) (x^{2} - 3)$ (b) $y = (x^2 + 1)\sqrt{x}$ .

Solution

$y^{'} = 2 (x^{2} - 3) + (2 x + 1) (2 x) = 6 x^{2} + 2 x - 6$ . Check by expanding: $y = 2 x^{3} + x^{2} - 6 x - 3$ , $y^{'} = 6 x^{2} + 2 x - 6$ ✓.
$y' = 2x\sqrt{x} + (x^2 + 1) \cdot \frac{1}{2\sqrt{x}} = \frac{4x^2 + x^2 + 1}{2\sqrt{x}} = \frac{5x^2 + 1}{2\sqrt{x}}$ , for $x \gt 0$ .

Important

Quotient rule. Where $v \ne 0$ ,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} .

Reason. Let $q = \frac{u}{v}$ , so $u = q v$ . The product rule gives $u^{'} = q^{'} v + q v^{'}$ , so $q' = \frac{u' - qv'}{v} = \frac{u' - \frac{u}{v}v'}{v} = \frac{u'v - uv'}{v^2}$ .

Worked example 2: quotients

Differentiate (a) $y = \frac{2x + 1}{x - 3}$ (b) $y = \frac{x^2}{x^2 + 1}$ .

Solution

$y' = \frac{2(x - 3) - (2x + 1)(1)}{(x - 3)^2} = \frac{-7}{(x - 3)^2}$ , for $x \ne 3$ .
$y' = \frac{2x(x^2 + 1) - x^2 \cdot 2x}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2}$ .

Warning

The order in the quotient rule matters: $u^{'} v - u v^{'}$ , "derivative of the top times the bottom, minus the top times the derivative of the bottom". Swapping the terms changes the sign of the answer. And the denominator is $v^{2}$ , not $v^{'}$ .

Important

Chain rule. If $y = f (u)$ and $u = g (x)$ are differentiable, then $y = f (g (x))$ is differentiable and

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, \qquad\text{that is}\qquad (f \circ g)'(x) = f'(g(x)) \cdot g'(x) .

"Differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inside." Special cases: $(u^n)' = nu^{n-1}u'$ and $(\sqrt{u})' = \frac{u'}{2\sqrt{u}}$ .

Why it is plausible: if $u$ changes 3 times as fast as $x$ , and $y$ changes 2 times as fast as $u$ , then $y$ changes $2 \times 3 = 6$ times as fast as $x$ .

Worked example 3: composites

Differentiate (a) $y = (3 x^{2} - 5)^{4}$ (b) $y = \sqrt{x^2 + 9}$ (c) $y = \frac{1}{(2x - 1)^3}$ .

Solution

Inside $u = 3 x^{2} - 5$ , $u^{'} = 6 x$ : $y' = 4(3x^2 - 5)^3 \cdot 6x = 24x(3x^2 - 5)^3$ .
$y' = \frac{2x}{2\sqrt{x^2 + 9}} = \frac{x}{\sqrt{x^2 + 9}}$ .
$y = (2x - 1)^{-3}$ : $y' = -3(2x - 1)^{-4} \cdot 2 = -\frac{6}{(2x - 1)^4}$ , for $x \ne \frac{1}{2}$ .

Worked example 4: rules together

Differentiate $y = x\sqrt{1 - x^2}$ for $-1 \lt x \lt 1$ , and simplify.

Solution Product rule, with the chain rule inside:

y' = 1 \cdot \sqrt{1 - x^2} + x \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \frac{(1 - x^2) - x^2}{\sqrt{1 - x^2}} = \frac{1 - 2x^2}{\sqrt{1 - x^2}} .

Definition

Higher-order derivatives: the derivative of $f^{'}$ is the second derivative $f^{''}$ (also $y^{''}$ or $\frac{d^2y}{dx^2}$ ); then $f^{'''}$ , and in general the $n$ th derivative $f^{(n)}$ .

Example: $f (x) = x^{4} - 2 x^{3} + 5$ : $f^{'} = 4 x^{3} - 6 x^{2}$ , $f^{''} = 12 x^{2} - 12 x$ , $f^{'''} = 24 x - 12$ , $f^{(4)} = 24$ , and $f^{(5)} = 0$ . A polynomial of degree $n$ has $f^{(n+1)} = 0$ .

Worked example 5: a pattern in the $n$ th derivative

Find $f^{(n)}(x)$ for $f(x) = \frac{1}{x}$ .

Solution $f' = -x^{-2}$ , $f'' = 2x^{-3}$ , $f''' = -6x^{-4}$ , $f^{(4)} = 24x^{-5}$ . The signs alternate and the coefficients are $1, 2, 6, 24 = n!$ , so

f^{(n)}(x) = \frac{(-1)^n\, n!}{x^{n+1}} .

(The pattern can be proved by mathematical induction: differentiating $(-1)^n n!\,x^{-(n+1)}$ gives $(-1)^{n+1}(n + 1)!\,x^{-(n+2)}$ .)

Tip

Before differentiating, choose the easiest route. $\frac{x^3 + 1}{x}$ is easier as $x^2 + x^{-1}$ than with the quotient rule; $\frac{5}{(x^2 + 1)^2}$ is easier as $5(x^2 + 1)^{-2}$ with the chain rule. Keep the quotient rule for genuine quotients.

3Blue1Brown: Visualizing the chain rule and product rule (about 17 min). Watch the product rule appear as the growth of a rectangle's area, and the chain rule as tiny nudges passed from one function to the next.

✦

Practice

Simplify each answer as far as is reasonable.

$y = (x^{3} - 2) (4 x + 5)$
$y = \frac{3x - 2}{x^2 + 1}$
$y = (5 - 2 x)^{6}$
$y = \sqrt{3x^2 - 2x + 7}$
$y = \frac{x}{\sqrt{x + 1}}$
$y = (x + 1)^{2} (x - 2)^{3}$ ; factorise your answer.
For $f(x) = \sqrt{x}$ , find $f^{''} (4)$ .
Find the $n$ th derivative of $f(x) = \frac{1}{1 - x}$ .

Show answersHide answers

$3 x^{2} (4 x + 5) + 4 (x^{3} - 2) = 16 x^{3} + 15 x^{2} - 8$ .
$\frac{3(x^2 + 1) - (3x - 2)2x}{(x^2 + 1)^2} = \frac{-3x^2 + 4x + 3}{(x^2 + 1)^2}$ .
$6 (5 - 2 x)^{5} (- 2) = - 12 (5 - 2 x)^{5}$ .
$\frac{6x - 2}{2\sqrt{3x^2 - 2x + 7}} = \frac{3x - 1}{\sqrt{3x^2 - 2x + 7}}$ .
$\frac{\sqrt{x + 1} - \frac{x}{2\sqrt{x + 1}}}{x + 1} = \frac{x + 2}{2(x + 1)^{\frac{3}{2}}}$ .
$2 (x + 1) (x - 2)^{3} + 3 (x + 1)^{2} (x - 2)^{2} = (x + 1) (x - 2)^{2} (5 x - 1)$ .
$f'' = -\frac{1}{4}x^{-\frac{3}{2}}$ , so $f''(4) = -\frac{1}{4} \times \frac{1}{8} = -\frac{1}{32}$ .
$f' = (1 - x)^{-2}$ , $f'' = 2(1 - x)^{-3}$ , …, $f^{(n)}(x) = \frac{n!}{(1 - x)^{n+1}}$ .

Quick check

Match each rule with its formula.

Tap one on the left, then its partner on the right.

Key points

Product $(u v)^{'} = u^{'} v + u v^{'}$ ; quotient $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ .
Chain rule: $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ ; in particular $(u^n)' = nu^{n-1}u'$ and $(\sqrt{u})' = \frac{u'}{2\sqrt{u}}$ .
Higher derivatives $f'', f''', \ldots, f^{(n)}$ : differentiate again; look for a pattern for $f^{(n)}$ .
Choose the easiest route: simplify or rewrite as powers before differentiating.

Sources and credits

Written by Ganzaa: the explanations, the diagram, 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: Leibniz, Nova Methodus pro Maximis et Minimis, Acta Eruditorum, October 1684, Wikimedia Commons, public domain.
Video: Visualizing the chain rule and product rule | Chapter 4, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1Rules that do the limits for you

Rules that do the limits for you