Sec 4 · Additional Mathematics

Differentiation

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1. Differentiating powers of x

Rule	Formula	Example
power	$\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$	$\dfrac{d}{dx}(x^{5}) = 5x^{4}$
constant	$\dfrac{d}{dx}(k) = 0$	$\dfrac{d}{dx}(7) = 0$
multiple	$\dfrac{d}{dx}(kx^{n}) = knx^{n - 1}$	$\dfrac{d}{dx}(3x^{4}) = 12x^{3}$
sum	differentiate term by term	$\dfrac{d}{dx}(x^{2} + 4x) = 2x + 4$

Write every term as a power of x first

\displaystyle y = 3x^{4} - \dfrac{2}{x^{2}} + 5\sqrt{x}

$y = 3x^{4} - 2x^{-2} + 5x^{\frac{1}{2}}$
$\dfrac{dy}{dx} = 12x^{3} + 4x^{-3} + \dfrac{5}{2}x^{-\frac{1}{2}}$
$\dfrac{dy}{dx} = 12x^{3} + \dfrac{4}{x^{3}} + \dfrac{5}{2\sqrt{x}}$

📖 Worked example

Given that $y = 3x^{4} - \dfrac{2}{x^{2}} + 5\sqrt{x}$ , find $\dfrac{dy}{dx}$ .

Step 1 of 3

Write every term as a power of x: 1 over xⁿ is x to the power −n, and a root is a fractional power

y = 3x^{4} - 2x^{-2} + 5x^{\frac{1}{2}}

Practise this →

2. The chain rule

Rule	Formula	Example
chain rule	$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$
a power of $ax + b$	$\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n - 1}$	$\dfrac{d}{dx}(3x - 1)^{4} = 12(3x - 1)^{3}$

Let u = the bracket

\displaystyle y = \dfrac{4}{(2x + 1)^{2}}

$u = 2x + 1$
$y = 4u^{-2}$
$\dfrac{du}{dx} = 2$
$\dfrac{dy}{du} = -8u^{-3}$
$\dfrac{dy}{dx} = -\dfrac{16}{(2x + 1)^{3}}$

📖 Worked example

Given that $y = \dfrac{3}{(3x - 3)^{3}}$ , find $\dfrac{dy}{dx}$ .

Step 1 of 5

Write y as a power of the bracket: 1 over a power is a negative power, and a square root is the power ½

y = 3(3x - 3)^{-3}

Practise this →

3. The product rule

Rule	Formula
product rule	$\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

u = x², v = (3x − 1)⁴

\displaystyle y = x^{2}(3x - 1)^{4}

$\dfrac{dy}{dx} = x^{2} \cdot 12(3x - 1)^{3} + (3x - 1)^{4} \cdot 2x$
$\dfrac{dy}{dx} = 2x(3x - 1)^{3}(9x - 1)$

📖 Worked example

Given that $y = x\sqrt{2x + 1}$ , find $\dfrac{dy}{dx}$ , giving your answer in factorised form.

Step 1 of 4

Let u be the first factor and v the second

u = x

v = \sqrt{2x + 1}

Practise this →

4. The quotient rule

Rule	Formula
quotient rule	$\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$

u = 2x + 1, v = x + 3

\displaystyle y = \dfrac{2x + 1}{x + 3}

$\dfrac{dy}{dx} = \dfrac{(x + 3)(2) - (2x + 1)(1)}{(x + 3)^{2}}$
$\dfrac{dy}{dx} = \dfrac{5}{(x + 3)^{2}}$

📖 Worked example

Given that $y = \dfrac{2x + 1}{\sqrt{x - 2}}$ , find $\dfrac{dy}{dx}$ , simplifying your answer.

Step 1 of 4

Let u be the top and v the bottom

u = 2x + 1

v = \sqrt{x - 2}

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5. Gradients and second derivatives

Name	Notation	Example: $y = x^{3} - 2x^{2}$
gradient function	$\dfrac{dy}{dx}$	$3x^{2} - 4x$
second derivative	$\dfrac{d^{2}y}{dx^{2}}$	$6x - 4$

Gradient of y = x³ − 2x² at x = 2

\displaystyle \dfrac{dy}{dx} = 3x^{2} - 4x

e.g.

\displaystyle 3(2)^{2} - 4(2)

\displaystyle = 4

📖 Worked example

Find the $x$ -coordinates of the points on the curve $y = 2x^{3} - 18x^{2} + 48x + 1$ at which the gradient is $0$ .

Step 1 of 4

Differentiate each term: bring the power down, take 1 off the power

\dfrac{dy}{dx} = 2(3)x^{2} - 18(2)x + 48(1)

= 6x^{2} - 36x + 48

Practise this →

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Sec 4 · Additional Mathematics

Differentiation

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1. Differentiating powers of x

Rule	Formula	Example
power	$\dfrac{d}{dx}(x^{n}) = nx^{n - 1}$	$\dfrac{d}{dx}(x^{5}) = 5x^{4}$
constant	$\dfrac{d}{dx}(k) = 0$	$\dfrac{d}{dx}(7) = 0$
multiple	$\dfrac{d}{dx}(kx^{n}) = knx^{n - 1}$	$\dfrac{d}{dx}(3x^{4}) = 12x^{3}$
sum	differentiate term by term	$\dfrac{d}{dx}(x^{2} + 4x) = 2x + 4$

Write every term as a power of x first

\displaystyle y = 3x^{4} - \dfrac{2}{x^{2}} + 5\sqrt{x}

$y = 3x^{4} - 2x^{-2} + 5x^{\frac{1}{2}}$
$\dfrac{dy}{dx} = 12x^{3} + 4x^{-3} + \dfrac{5}{2}x^{-\frac{1}{2}}$
$\dfrac{dy}{dx} = 12x^{3} + \dfrac{4}{x^{3}} + \dfrac{5}{2\sqrt{x}}$

📖 Worked example

Given that $y = 3x^{4} - \dfrac{2}{x^{2}} + 5\sqrt{x}$ , find $\dfrac{dy}{dx}$ .

Step 1 of 3

Write every term as a power of x: 1 over xⁿ is x to the power −n, and a root is a fractional power

y = 3x^{4} - 2x^{-2} + 5x^{\frac{1}{2}}

Practise this →

2. The chain rule

Rule	Formula	Example
chain rule	$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$
a power of $ax + b$	$\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n - 1}$	$\dfrac{d}{dx}(3x - 1)^{4} = 12(3x - 1)^{3}$

Let u = the bracket

\displaystyle y = \dfrac{4}{(2x + 1)^{2}}

$u = 2x + 1$
$y = 4u^{-2}$
$\dfrac{du}{dx} = 2$
$\dfrac{dy}{du} = -8u^{-3}$
$\dfrac{dy}{dx} = -\dfrac{16}{(2x + 1)^{3}}$

📖 Worked example

Given that $y = \dfrac{3}{(3x - 3)^{3}}$ , find $\dfrac{dy}{dx}$ .

Step 1 of 5

Write y as a power of the bracket: 1 over a power is a negative power, and a square root is the power ½

y = 3(3x - 3)^{-3}

Practise this →

3. The product rule

Rule	Formula
product rule	$\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

u = x², v = (3x − 1)⁴

\displaystyle y = x^{2}(3x - 1)^{4}

$\dfrac{dy}{dx} = x^{2} \cdot 12(3x - 1)^{3} + (3x - 1)^{4} \cdot 2x$
$\dfrac{dy}{dx} = 2x(3x - 1)^{3}(9x - 1)$

📖 Worked example

Given that $y = x\sqrt{2x + 1}$ , find $\dfrac{dy}{dx}$ , giving your answer in factorised form.

Step 1 of 4

Let u be the first factor and v the second

u = x

v = \sqrt{2x + 1}

Practise this →

4. The quotient rule

Rule	Formula
quotient rule	$\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}$

u = 2x + 1, v = x + 3

\displaystyle y = \dfrac{2x + 1}{x + 3}

$\dfrac{dy}{dx} = \dfrac{(x + 3)(2) - (2x + 1)(1)}{(x + 3)^{2}}$
$\dfrac{dy}{dx} = \dfrac{5}{(x + 3)^{2}}$

📖 Worked example

Given that $y = \dfrac{2x + 1}{\sqrt{x - 2}}$ , find $\dfrac{dy}{dx}$ , simplifying your answer.

Step 1 of 4

Let u be the top and v the bottom

u = 2x + 1

v = \sqrt{x - 2}

Practise this →

5. Gradients and second derivatives

Name	Notation	Example: $y = x^{3} - 2x^{2}$
gradient function	$\dfrac{dy}{dx}$	$3x^{2} - 4x$
second derivative	$\dfrac{d^{2}y}{dx^{2}}$	$6x - 4$

Gradient of y = x³ − 2x² at x = 2

\displaystyle \dfrac{dy}{dx} = 3x^{2} - 4x

e.g.

\displaystyle 3(2)^{2} - 4(2)

\displaystyle = 4

📖 Worked example

Find the $x$ -coordinates of the points on the curve $y = 2x^{3} - 18x^{2} + 48x + 1$ at which the gradient is $0$ .

Step 1 of 4

Differentiate each term: bring the power down, take 1 off the power

\dfrac{dy}{dx} = 2(3)x^{2} - 18(2)x + 48(1)

= 6x^{2} - 36x + 48

Practise this →

Try a question: Differentiation

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