Chapter 11 · Additional Maths · Differentiation techniques

Differentiation rules: power, chain, product and quotient

The derivative tells you how steeply a curve climbs at any point. This chapter builds the rules that find it fast: the power rule for any rational index, the chain rule for a bracket raised to a power, and the product and quotient rules for functions multiplied or divided. You will also meet the second derivative and use f'(a) as the gradient of a tangent. Derivatives of sin, cos, tan, e^x and ln x use the same rules and come in chapter 13.

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Chapter
11

Figure 1 · a tangent line touching a curve: the gradient of the curve at that point is the gradient of the line (y = x^3 - 3x at the origin)

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Power rule

Multiply by the power, then lower the power by 1. Write roots and reciprocals as powers first: $\sqrt{x} = x^{\frac{1}{2}}$ , $\dfrac{1}{x^{2}} = x^{-2}$ . A constant term differentiates to 0.

Chain rule

$\dfrac{d}{dx}f(g(x)) = f'(g(x)) \times g'(x)$ : outside derivative, bracket kept, then times the derivative of the bracket. For $(ax + b)^{n}$ the answer is $na(ax + b)^{n-1}$ .

Product and quotient

$(uv)' = u'v + uv'$ and $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^{2}}$ . Mind the minus sign and the order in the quotient rule; a one-term denominator is easier as a negative power.

By the end of this chapter

Read dy/dx and f'(x) as the gradient of the tangent at a point and as a rate of change, and use the notation dy/dx, f'(x), d²y/dx² and f''(x).
Differentiate any sum of powers of x, including negative and fractional indices, after rewriting roots and reciprocals in index form.
Differentiate a power of a bracket, k(ax + b)^n and k(g(x))^n, with the chain rule.
Differentiate products and quotients, choose the efficient method, and simplify or factorise the answer.
Find and interpret the second derivative, and show that a relation between y, dy/dx and d²y/dx² holds.
Use the gradient at a point to find the equation of a tangent and a normal, a point with a given gradient, or an unknown constant.

✶

Start here

Maths In Real Life

Two situations where a gradient or a rate of change decides what can be built or booked. The pictures come first; the working in each solution comes from the same numbers.

$A smooth concave curved concrete access ramp seen from the side, rising gently from level ground at the left to a steeper top end at the right. The profile of the ramp is traced as a curve, the point P and the top end T are marked, a dashed straight line touches the ramp at P, and the horizontal length of the ramp is measured$

y = \dfrac{x^{2}}{160}

12\text{ m}

P

T

\text{tangent at } P

\text{vertical scale exaggerated}

Maths In Real Life 1

A curved access ramp

A ramp that is curved instead of straight is gentle at the start and steeper towards the top. Because its steepness changes from place to place, the gradient at one point has to be found by differentiating.

A community centre is building a curved access ramp. At a horizontal distance of $x$ metres from the start of the ramp, the surface of the ramp is $y$ metres above the ground, where $y = \frac{x^{2}}{160}$ , from the start of the ramp up to $x = 12$ .

(a)The designer wants the gradient to be at most $\frac{1}{10}$ . Find the value of $x$ at which the gradient of the ramp first reaches this limit, and the height of the ramp there.
(b)Find the gradient of the ramp at its top end, where $x = 12$ . Does the top end of the ramp meet the designer's limit? Give a reason.

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Colour keyderivative of a term or factorfinal derivative

(a)

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{\frac{1}{80}x}

Multiply by the power, then lower the power by

1

\displaystyle \frac{x}{80} = \frac{1}{10}

The gradient is given, so set the derivative equal to it.

\displaystyle \frac{x}{80} - \frac{1}{10} = 0

Move every term to the left.

\displaystyle \frac{x - 8}{80} = 0 \iff \textcolor{#010106}{x = 8}

A fraction is zero only when its top is zero, so

x - 8 = 0

\displaystyle y\;= \frac{\left(8\right)^{2}}{160} = \textcolor{#010105}{\frac{2}{5}}

Put

x = 8

into the curve.

(b)

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{\frac{1}{80}x}

Multiply by the power, then lower the power by

1

\displaystyle m\;= \frac{\left(12\right)}{80} = \textcolor{#010106}{\frac{3}{20}}

Put

x = 12

into the derivative to get the gradient of the tangent there.

Compare with the designer's limit:

\frac{3}{20} = 0.15

and

\frac{1}{10} = 0.1

, so

\frac{3}{20} > \frac{1}{10}

. The top end of the ramp is steeper than the limit allows, so this section does not meet the designer's rule.

Answer

(a)

x = 8

m, where the ramp is

0.4

m high (

y = \frac{2}{5}

)

(b)

m = \frac{3}{20}

. This is more than

\frac{1}{10}

, so the top end does not meet the designer's limit.

$A plain side view of an empty white chartered coach with a row of rectangular windows and black wheels, parked on a grey road at a school gate under a pale sky.$

x \text{ passengers}

\text{side view}

Maths In Real Life 2

The cost per head of a chartered coach

A coach company charges one fixed fee for the vehicle and driver, however many people ride, plus a smaller charge for each passenger. Dividing the total by the number of passengers gives the cost per head, so the more people book, the less each one pays. A quotient shows exactly how quickly that falls.

A class hires a coach for a trip. When $x$ passengers book, the cost per head, $y$ dollars, is modelled by $y = \dfrac{ 12 x + 360 }{ x }$ , for $x > 0$ . The top of the fraction is the total cost of the trip in dollars: a fixed charge for the coach plus a charge for each passenger.

(a)Use the quotient rule to find $\dfrac{dy}{dx}$ . Check your answer by writing $y$ as a sum of powers of $x$ and differentiating again.
(b)Find the rate at which the cost per head is changing when $20$ passengers have booked. Give your answer in dollars per passenger.

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Colour keyderivative of a term or factorfinal derivative

(a)

Differentiate the top and the bottom separately.

\displaystyle \frac{d}{dx}\left(12 x + 360\right) = \textcolor{#010105}{12}

Power rule on each term.

\displaystyle \frac{d}{dx}\left(x\right) = \textcolor{#010105}{1}

Power rule on each term.

\displaystyle \frac{dy}{dx}\;= \dfrac{12 x - \left(12 x + 360\right)}{x^{2}}

Quotient rule: (derivative of the top times the bottom) minus (the top times the derivative of the bottom), over the bottom squared.

\displaystyle = \dfrac{\textcolor{#010106}{-360}}{x^{2}}

Expand the top and collect like terms. The top is a single number, so nothing cancels with the bottom.

Check by a second route: divide first, so that the quotient becomes a sum of powers of

x

\displaystyle y\;= 12 + 360x^{-1}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(12\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{d}{dx}\left(360x^{-1}\right) = \textcolor{#010105}{-360x^{-2}}

Multiply the coefficient

360

by the power

-1

to get

-360

, then lower the power to

-2

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{-360x^{-2}}

Add the derivatives of the terms.

\displaystyle = - \frac{360}{x^{2}}

Write the negative and fractional powers as fractions and roots again.

The two routes give the same derivative.

(b)

\displaystyle y\;= 12 + 360x^{-1}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(12\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{d}{dx}\left(360x^{-1}\right) = \textcolor{#010105}{-360x^{-2}}

Multiply the coefficient

360

by the power

-1

to get

-360

, then lower the power to

-2

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{-360x^{-2}}

Add the derivatives of the terms.

\displaystyle = - \frac{360}{x^{2}}

Write the negative and fractional powers as fractions and roots again.

\displaystyle m\;= - \frac{360}{\left(20\right)^{2}} = \textcolor{#010106}{- \frac{9}{10}}

Put

x = 20

into the derivative. The gradient

\dfrac{dy}{dx}

there is the rate of change of

y

with respect to

x

Answer

(a)

\frac{dy}{dx} = - \frac{360}{x^{2}}

(b)

m = -0.9

dollars per extra passenger: the cost per head is falling

Supports K341-C1-01 · K341-C1-02 · K341-C1-03

The derivative as the gradient of a tangent

A straight line has the same steepness everywhere, but a curve does not: it is steep in one place and nearly flat in another. The gradient of a curve at a point means the gradient of the tangent there, the straight line that just touches the curve at that point and runs in the same direction. This section shows where that idea comes from and what the notation means. The next sections build quick rules for finding the gradient.

P

is a fixed point on the curve. The dashed chords from

P

Q

and then to

R

lean closer and closer to the solid tangent at

P

as the second point moves towards

P

Here is the idea behind the tangent. Join $P$ to a second point on the curve by a chord, and work out the gradient of that chord. Now slide the second point along the curve towards $P$ . The chord swings round and its gradient settles down to one value, and that value is the gradient of the tangent at $P$ . Doing this with algebra for a general point gives the derivative. You are not asked to carry out this limit process in the exam; you only need to know that the derivative is the gradient of the tangent.

Think About It

The curve is

y = x^{2}

and

P

is the point where

x = 1

. Let

Q

be the point of the curve where

x = 1 + h

, with

h

a small positive number. What is the gradient of the chord

PQ

, and what happens to it as

h

gets smaller?

Show the full workingHide the working

\displaystyle m\;= \dfrac{\left(1 + h\right)^{2} - 1}{h}

The rise is the difference of the two

y

-values and the run is

h

\displaystyle = \dfrac{2h + h^{2}}{h} = 2 + h

Expand the bracket, then divide the top and the bottom by

h

, which is not zero.

h

gets smaller,

2 + h

gets closer and closer to

2

, so the chords settle onto a line of gradient

2

Answer

The chord gradient is

2 + h

, which approaches

2

: the gradient of the tangent to

y = x^{2}

x = 1

The Core Idea

The derivative is the gradient of the tangent.

For a curve $y = f(x)$ the derivative is written $\dfrac{dy}{dx}$ or $f'(x)$ . Its value at $x = a$ is the gradient of the tangent at the point where $x = a$ , written $f'(a)$ . Because the gradient says how fast $y$ changes as $x$ increases, it is also the rate of change of $y$ with respect to $x$ , measured in units of $y$ per unit of $x$ .

For the next few Examples you need one rule before it is explained properly in the next section: to differentiate a term $ax^{n}$ , multiply by the power and lower the power by 1, so $ax^{n}$ becomes $an{}x^{n-1}$ . A number on its own has derivative $0$ , and the terms of a sum are differentiated one at a time.

SGSchoolKaki Tip

Always keep the two ideas apart. $f'(x)$ is a function of $x$ , the general gradient formula. $f'(a)$ is a number, the gradient at one particular point. You find the formula first, then put in the value of $x$ .

Example 1.1

The derivative, then the gradient at a point

A curve has equation $y = x^{2} + 3 x$ . Find $\dfrac{dy}{dx}$ and the gradient of the curve at the point where $x = 2$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(3x\right) = \textcolor{#010105}{3}

The power is

1

: multiply

3

1

, and

x^{0}

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x + 3}

Add the derivatives of the terms.

\displaystyle m\;= 2 \times \left(2\right) + 3 = \textcolor{#010106}{7}

Put

x = 2

into the derivative to get the gradient of the tangent there.

Answer

\frac{dy}{dx} = 2 x + 3

;

m = 7

Your Turn 1.1

A curve has equation $y = x^{2} - 5 x$ . Find $\dfrac{dy}{dx}$ and the gradient of the curve at the point where $x = 4$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(-5x\right) = \textcolor{#010105}{-5}

The power is

1

: multiply

-5

1

, and

x^{0}

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x - 5}

Add the derivatives of the terms.

\displaystyle m\;= 2 \times \left(4\right) - 5 = \textcolor{#010106}{3}

Put

x = 4

into the derivative to get the gradient of the tangent there.

Answer

\frac{dy}{dx} = 2 x - 5

;

m = 3

Example 1.2

Function notation: f'(x) and f'(a)

Given that $f(x) = x^{3} - 3 x^{2}$ , find $f'(x)$ and the value of $f'(-1)$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{3}\right) = \textcolor{#010105}{3x^{2}}

The coefficient is

1

: bring the power

3

down and lower the power to

2

\displaystyle \frac{d}{dx}\left(-3x^{2}\right) = \textcolor{#010105}{-6x}

Multiply the coefficient

-3

by the power

2

to get

-6

, then lower the power to

1

\displaystyle f'(x)\;= \textcolor{#010106}{3x^{2} - 6x}

Add the derivatives of the terms.

\displaystyle m\;= 3 \times \left(-1\right)^{2} - 6 \times \left(-1\right) = \textcolor{#010106}{9}

Put

x = -1

into the derivative to get the gradient of the tangent there.

Answer

f'(x) = 3 x^{2} - 6 x

;

m = 9

Your Turn 1.2

Given that $f(x) = x^{3} + 2 x^{2}$ , find $f'(x)$ and the value of $f'(-2)$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{3}\right) = \textcolor{#010105}{3x^{2}}

The coefficient is

1

: bring the power

3

down and lower the power to

2

\displaystyle \frac{d}{dx}\left(2x^{2}\right) = \textcolor{#010105}{4x}

Multiply the coefficient

2

by the power

2

to get

4

, then lower the power to

1

\displaystyle f'(x)\;= \textcolor{#010106}{3x^{2} + 4x}

Add the derivatives of the terms.

\displaystyle m\;= 3 \times \left(-2\right)^{2} + 4 \times \left(-2\right) = \textcolor{#010106}{4}

Put

x = -2

into the derivative to get the gradient of the tangent there.

Answer

f'(x) = 3 x^{2} + 4 x

;

m = 4

f'(-1)

means put

x = -1

into the derivative

f'(x)

. It does not mean the derivative of

f(-1)

: that is a single number, and the derivative of a number is

0

. Differentiate first, substitute second.

Example 1.3

The gradient as a rate of change

The curve $y = 5 x^{2} - 2 x$ describes how a quantity $y$ changes as $x$ changes. Find $\dfrac{dy}{dx}$ , and the rate of change of $y$ with respect to $x$ when $x = 3$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(5x^{2}\right) = \textcolor{#010105}{10x}

Multiply the coefficient

5

by the power

2

to get

10

, then lower the power to

1

\displaystyle \frac{d}{dx}\left(-2x\right) = \textcolor{#010105}{-2}

The power is

1

: multiply

-2

1

, and

x^{0}

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{10x - 2}

Add the derivatives of the terms.

\displaystyle m\;= 10 \times \left(3\right) - 2 = \textcolor{#010106}{28}

Put

x = 3

into the derivative. The gradient

\dfrac{dy}{dx}

there is the rate of change of

y

with respect to

x

Answer

\frac{dy}{dx} = 10 x - 2

;

m = 28

units of

y

per unit of

x

Your Turn 1.3

The curve $y = 3 x^{2} + 4 x$ describes how a quantity $y$ changes as $x$ changes. Find $\dfrac{dy}{dx}$ , and the rate of change of $y$ with respect to $x$ when $x = 2$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(3x^{2}\right) = \textcolor{#010105}{6x}

Multiply the coefficient

3

by the power

2

to get

6

, then lower the power to

1

\displaystyle \frac{d}{dx}\left(4x\right) = \textcolor{#010105}{4}

The power is

1

: multiply

4

1

, and

x^{0}

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{6x + 4}

Add the derivatives of the terms.

\displaystyle m\;= 6 \times \left(2\right) + 4 = \textcolor{#010106}{16}

Put

x = 2

into the derivative. The gradient

\dfrac{dy}{dx}

there is the rate of change of

y

with respect to

x

Answer

\frac{dy}{dx} = 6 x + 4

;

m = 16

units of

y

per unit of

x

Example 1.4

Working backwards: where is the gradient given?

Find the coordinates of the point on the curve $y = x^{2} - 6 x + 5$ at which the gradient of the curve is $2$ .

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(-6x\right) = \textcolor{#010105}{-6}

The power is

1

: multiply

-6

1

, and

x^{0}

1

\displaystyle \frac{d}{dx}\left(5\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x - 6}

Add the derivatives of the terms.

\displaystyle 2 x - 6 = 2

The gradient is given, so set the derivative equal to it.

\displaystyle 2 x - 8 = 0

Move every term to the left.

\displaystyle 2 \left(x - 4\right) = 0 \iff \textcolor{#010106}{x = 4}

Each factor can be zero.

\displaystyle y\;= \left(4\right)^{2} - 6 \times \left(4\right) + 5 = \textcolor{#010105}{-3}

Put

x = 4

into the curve.

Answer

(4, -3)

Your Turn 1.4

Find the coordinates of the point on the curve $y = x^{2} + 4 x - 1$ at which the gradient of the curve is $-2$ .

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(4x\right) = \textcolor{#010105}{4}

The power is

1

: multiply

4

1

, and

x^{0}

1

\displaystyle \frac{d}{dx}\left(-1\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x + 4}

Add the derivatives of the terms.

\displaystyle 2 x + 4 = -2

The gradient is given, so set the derivative equal to it.

\displaystyle 2 x + 6 = 0

Move every term to the left.

\displaystyle 2 \left(x + 3\right) = 0 \iff \textcolor{#010106}{x = -3}

Each factor can be zero.

\displaystyle y\;= \left(-3\right)^{2} + 4 \times \left(-3\right) - 1 = \textcolor{#010105}{-4}

Put

x = -3

into the curve.

Answer

(-3, -4)

Differentiating the derivative gives the second derivative. It is written $\dfrac{d^{2}y}{dx^{2}}$ or $f''(x)$ : the derivative of $\dfrac{dy}{dx}$ . It tells you how the gradient itself is changing. This chapter only computes it and reads it as the rate at which the gradient changes.

\dfrac{d^{2}y}{dx^{2}}

is a single symbol for the second derivative. It is not

\left(\dfrac{dy}{dx}\right)^{2}

, which would mean the first derivative multiplied by itself.

Example 1.5

dy/dx and d2y/dx2 of the same function

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

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{5}\right) = \textcolor{#010105}{5x^{4}}

The coefficient is

1

: bring the power

5

down and lower the power to

4

\displaystyle \frac{d}{dx}\left(-4x^{2}\right) = \textcolor{#010105}{-8x}

Multiply the coefficient

-4

by the power

2

to get

-8

, then lower the power to

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{5x^{4} - 8x}

Add the derivatives of the terms.

Differentiate the first derivative once more.

\displaystyle \frac{d}{dx}\left(5x^{4}\right) = \textcolor{#010105}{20x^{3}}

Multiply by the power and lower the power by

1

\displaystyle \frac{d}{dx}\left(-8x\right) = \textcolor{#010105}{-8}

Multiply by the power and lower the power by

1

\displaystyle \frac{d^{2}y}{dx^{2}}\;= \textcolor{#010106}{20x^{3} - 8}

Add the results.

\displaystyle = 20 x^{3} - 8

Write the powers as fractions and roots again.

Answer

\frac{dy}{dx} = 5 x^{4} - 8 x

;

\frac{d^{2}y}{dx^{2}} = 20 x^{3} - 8

Your Turn 1.5

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

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Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(x^{4}\right) = \textcolor{#010105}{4x^{3}}

The coefficient is

1

: bring the power

4

down and lower the power to

3

\displaystyle \frac{d}{dx}\left(3x^{3}\right) = \textcolor{#010105}{9x^{2}}

Multiply the coefficient

3

by the power

3

to get

9

, then lower the power to

2

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{4x^{3} + 9x^{2}}

Add the derivatives of the terms.

Differentiate the first derivative once more.

\displaystyle \frac{d}{dx}\left(4x^{3}\right) = \textcolor{#010105}{12x^{2}}

Multiply by the power and lower the power by

1

\displaystyle \frac{d}{dx}\left(9x^{2}\right) = \textcolor{#010105}{18x}

Multiply by the power and lower the power by

1

\displaystyle \frac{d^{2}y}{dx^{2}}\;= \textcolor{#010106}{12x^{2} + 18x}

Add the results.

\displaystyle = 12 x^{2} + 18 x

Write the powers as fractions and roots again.

Answer

\frac{dy}{dx} = 4 x^{3} + 9 x^{2}

;

\frac{d^{2}y}{dx^{2}} = 12 x^{2} + 18 x

Links To

A gradient is the rise divided by the run of a line, and a perpendicular line has gradient $-\dfrac{1}{m}$ . Both ideas are rebuilt in Chapter 7: Coordinate geometry: lines, areas and circles

Supports K341-C1-04 (x^n part)

The power rule, sums and differences

Almost every function in this chapter is built from powers of $x$ . One rule handles them all, and it works for whole-number, negative and fractional indices alike.

The Core Idea

Multiply by the power, then lower the power by 1.

For any rational number $n$ : $\dfrac{d}{dx}\left(x^{n}\right) = nx^{n-1}$ . A constant multiple stays in front, $\dfrac{d}{dx}\left(kx^{n}\right) = kn{}x^{n-1}$ , and a sum or difference is differentiated term by term. A term that is just a number differentiates to 0. When $n$ is a fraction with an even denominator (such as $\dfrac{1}{2}$ ), $x$ must be positive; when $n$ is negative, $x$ is not $0$ .

The rule needs every term written as a number times a power of $x$ . A root is a power with a fractional index and a reciprocal is a power with a negative index, so rewrite first, then differentiate, and write the answer back as a fraction or a root at the end.

Rewrite first, then differentiate
Form given	Index form	Derivative
$\dfrac{1}{x^{3}}$	$x^{-3}$	$-3x^{-4}$ , that is $-\dfrac{3}{x^{4}}$
$\sqrt{x}$	$x^{\frac{1}{2}}$	$\dfrac{1}{2}x^{-\frac{1}{2}}$ , that is $\dfrac{1}{2\sqrt{x}}$
$\sqrt[3]{x}$	$x^{\frac{1}{3}}$	$\dfrac{1}{3}x^{-\frac{2}{3}}$
$\dfrac{5}{x^{2}}$	$5x^{-2}$	$-10x^{-3}$ , that is $-\dfrac{10}{x^{3}}$

A number added or taken away: what happens to the gradient
Function	Gradient at x = 1	Picture
$y = x^{2}$	$2$	tangent where $x = 1$
$y = x^{2} + 2$	$2$ : unchanged by the $+ 2$	moved up, same steepness
$y = x^{2} - 1$	$2$ : unchanged by the $- 1$	moved down, same steepness

Moving a graph up or down does not change how steep it is, which is why a constant term differentiates to $0$ .

Example 2.1

A polynomial with a constant term

Find $\dfrac{dy}{dx}$ when $y = 4 x^{3} - 5 x^{2} + 7 x - 9$ .

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(4x^{3}\right) = \textcolor{#010105}{12x^{2}}

Multiply the coefficient

4

by the power

3

to get

12

, then lower the power to

2

\displaystyle \frac{d}{dx}\left(-5x^{2}\right) = \textcolor{#010105}{-10x}

Multiply the coefficient

-5

by the power

2

to get

-10

, then lower the power to

1

\displaystyle \frac{d}{dx}\left(7x\right) = \textcolor{#010105}{7}

The power is

1

: multiply

7

1

, and

x^{0}

1

\displaystyle \frac{d}{dx}\left(-9\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{12x^{2} - 10x + 7}

Add the derivatives of the terms.

Answer

\frac{dy}{dx} = 12 x^{2} - 10 x + 7

Your Turn 2.1

Find $\dfrac{dy}{dx}$ when $y = 6 x^{4} + 2 x^{3} - x + 8$ .

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(6x^{4}\right) = \textcolor{#010105}{24x^{3}}

Multiply the coefficient

6

by the power

4

to get

24

, then lower the power to

3

\displaystyle \frac{d}{dx}\left(2x^{3}\right) = \textcolor{#010105}{6x^{2}}

Multiply the coefficient

2

by the power

3

to get

6

, then lower the power to

2

\displaystyle \frac{d}{dx}\left(-x\right) = \textcolor{#010105}{-1}

The power is

1

: multiply

-1

1

, and

x^{0}

1

\displaystyle \frac{d}{dx}\left(8\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{24x^{3} + 6x^{2} - 1}

Add the derivatives of the terms.

Answer

\frac{dy}{dx} = 24 x^{3} + 6 x^{2} - 1

Example 2.2

A reciprocal term: a negative index

Differentiate $y = x^{2} + \frac{3}{x}$ with respect to $x$ .

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= x^{2} + 3x^{-1}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(3x^{-1}\right) = \textcolor{#010105}{-3x^{-2}}

Multiply the coefficient

3

by the power

-1

to get

-3

, then lower the power to

-2

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x - 3x^{-2}}

Add the derivatives of the terms.

\displaystyle = 2 x - \frac{3}{x^{2}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = 2 x - \frac{3}{x^{2}}

Your Turn 2.2

Differentiate $y = 2 x^{3} - \frac{4}{x^{2}}$ with respect to $x$ .

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= 2x^{3} - 4x^{-2}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(2x^{3}\right) = \textcolor{#010105}{6x^{2}}

Multiply the coefficient

2

by the power

3

to get

6

, then lower the power to

2

\displaystyle \frac{d}{dx}\left(-4x^{-2}\right) = \textcolor{#010105}{8x^{-3}}

Multiply the coefficient

-4

by the power

-2

to get

8

, then lower the power to

-3

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{6x^{2} + 8x^{-3}}

Add the derivatives of the terms.

\displaystyle = 6 x^{2} + \frac{8}{x^{3}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = 6 x^{2} + \frac{8}{x^{3}}

Think About It

What does the power rule give for

n = 0

, and does that agree with the rule that a constant has derivative

0

Show the full workingHide the working

Since

x^{0} = 1

, the function

x^{0}

is a constant.

Any non-zero number to the power

0

1

\displaystyle \dfrac{d}{dx}\left(x^{0}\right) = 0 \times x^{-1} = 0

The power rule with

n = 0

: the factor

0

in front kills the whole term.

Answer

The rule gives

0

, so it agrees with the fact that a constant has derivative

0

Example 2.3

A square root term: a fractional index

Find $\dfrac{dy}{dx}$ when $y = - 6 \sqrt{x} + x^{2}$ , for $x > 0$ .

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= x^{2} - 6x^{\frac{1}{2}}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(x^{2}\right) = \textcolor{#010105}{2x}

The coefficient is

1

: bring the power

2

down and lower the power to

1

\displaystyle \frac{d}{dx}\left(-6x^{\frac{1}{2}}\right) = \textcolor{#010105}{-3x^{- \frac{1}{2}}}

Multiply the coefficient

-6

by the power

\frac{1}{2}

to get

-3

, then lower the power to

- \frac{1}{2}

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2x - 3x^{- \frac{1}{2}}}

Add the derivatives of the terms.

\displaystyle = 2 x - \frac{3}{\sqrt{x}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = 2 x - \frac{3}{\sqrt{x}}

Your Turn 2.3

Find $\dfrac{dy}{dx}$ when $y = 8 \sqrt{x} + x^{3}$ , for $x > 0$ .

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= x^{3} + 8x^{\frac{1}{2}}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(x^{3}\right) = \textcolor{#010105}{3x^{2}}

The coefficient is

1

: bring the power

3

down and lower the power to

2

\displaystyle \frac{d}{dx}\left(8x^{\frac{1}{2}}\right) = \textcolor{#010105}{4x^{- \frac{1}{2}}}

Multiply the coefficient

8

by the power

\frac{1}{2}

to get

4

, then lower the power to

- \frac{1}{2}

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{3x^{2} + 4x^{- \frac{1}{2}}}

Add the derivatives of the terms.

\displaystyle = 3 x^{2} + \frac{4}{\sqrt{x}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = 3 x^{2} + \frac{4}{\sqrt{x}}

A fraction with a single term underneath is not a quotient problem. Split it into separate fractions, one for each term on top, and every piece becomes a number times a power of $x$ .

Example 2.4

A fraction split into terms

Find $\dfrac{dy}{dx}$ when $y = \frac{x^{2} + 5}{x^{2}}$ . Split the fraction into terms first.

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= 1 + 5x^{-2}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(1\right) = \textcolor{#010105}{0}

The derivative of a constant is

0

\displaystyle \frac{d}{dx}\left(5x^{-2}\right) = \textcolor{#010105}{-10x^{-3}}

Multiply the coefficient

5

by the power

-2

to get

-10

, then lower the power to

-3

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{-10x^{-3}}

Add the derivatives of the terms.

\displaystyle = - \frac{10}{x^{3}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = - \frac{10}{x^{3}}

Your Turn 2.4

Find $\dfrac{dy}{dx}$ when $y = \frac{2 x^{3} - 1}{x^{2}}$ . Split the fraction into terms first.

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle y\;= 2x - x^{-2}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{d}{dx}\left(2x\right) = \textcolor{#010105}{2}

The power is

1

: multiply

2

1

, and

x^{0}

1

\displaystyle \frac{d}{dx}\left(-x^{-2}\right) = \textcolor{#010105}{2x^{-3}}

Multiply the coefficient

-1

by the power

-2

to get

2

, then lower the power to

-3

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2 + 2x^{-3}}

Add the derivatives of the terms.

\displaystyle = 2 + \frac{2}{x^{3}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = 2 + \frac{2}{x^{3}}

Example 2.5

A constant taken out of the denominator

Differentiate $y = \frac{1}{4 x^{2}}$ with respect to $x$ .

Show solutionHide solution

Colour keyfinal derivative

\displaystyle y\;= \frac{1}{4}x^{-2}

Write every root and reciprocal as a power of

x

: a root has a fractional index and

x

in the denominator has a negative index.

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{-\frac{1}{2}x^{-3}}

Multiply by the power, then lower the power by

1

\displaystyle = - \frac{1}{2 x^{3}}

Write the negative and fractional powers as fractions and roots again.

Answer

\frac{dy}{dx} = - \frac{1}{2 x^{3}}

SGSchoolKaki Tip

A number in the denominator of a fraction is a constant multiplier. $\dfrac{1}{4x^{2}}$ is $\dfrac{1}{4}$ times $x^{-2}$ , so the $\dfrac{1}{4}$ stays in front and only $x^{-2}$ is differentiated.

A power of a bracket that holds only a multiple of

x

must be expanded before the power rule is used on it.

(2x)^{3}

2^{3}x^{3} = 8x^{3}

, not

2x^{3}

: the number inside the bracket is cubed too. The derivative of

(2x)^{3}

is therefore

24x^{2}

, not

6x^{2}

Example 2.6

Expanding a bracket of a single term

Find $\dfrac{dy}{dx}$ when $y = 8 x^{3} + x$ .

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(8x^{3}\right) = \textcolor{#010105}{24x^{2}}

Multiply the coefficient

8

by the power

3

to get

24

, then lower the power to

2

\displaystyle \frac{d}{dx}\left(x\right) = \textcolor{#010105}{1}

The power is

1

, so the term is

x

and its gradient is

1

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{24x^{2} + 1}

Add the derivatives of the terms.

Answer

\frac{dy}{dx} = 24 x^{2} + 1

Example 2.7

An unknown constant from a given gradient

The gradient of the curve $y = k x^{3} - 4 x$ , where $k$ is a constant, is $20$ at the point where $x = 2$ . Find the value of $k$ .

Show solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(kx^{3}\right) = \textcolor{#010105}{3 kx^{2}}

Power rule on the term; the unknown

k

is a constant multiplier.

\displaystyle \frac{d}{dx}\left(-4x\right) = \textcolor{#010105}{-4}

Power rule on the term; the unknown

k

is a constant multiplier.

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{3 kx^{2} - 4}

Add the derivatives of the terms.

\displaystyle 3 \times \left(2\right)^{2} \times k - 4 = 20

The gradient at

x = 2

is given, so substitute

x = 2

and set the result equal to

20

\displaystyle k\;= \textcolor{#010106}{2}

Solve the linear equation for

k

Answer

k = 2

Your Turn 2.7

The gradient of the curve $y = k x^{2} + 6 x$ , where $k$ is a constant, is $24$ at the point where $x = 3$ . Find the value of $k$ .

Show full solutionHide solution

Colour keyderivative of a term or factorfinal derivative

\displaystyle \frac{d}{dx}\left(kx^{2}\right) = \textcolor{#010105}{2 kx}

Power rule on the term; the unknown

k

is a constant multiplier.

\displaystyle \frac{d}{dx}\left(6x\right) = \textcolor{#010105}{6}

Power rule on the term; the unknown

k

is a constant multiplier.

\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{2 kx + 6}

Add the derivatives of the terms.

\displaystyle 2 \times \left(3\right) \times k + 6 = 24

The gradient at

x = 3

is given, so substitute

x = 3

and set the result equal to

24

\displaystyle k\;= \textcolor{#010106}{3}

Solve the linear equation for

k

Answer

k = 3

Supports K341-C1-06

The chain rule: powers of a bracket

The power rule needs a bare $x$ . A power of a bracket, such as $(2x + 5)^{4}$ , is different: the bracket is a function of $x$ in its own right. You could multiply it out, but that is slow and a single slip ruins the answer. The chain rule differentiates it directly.

The Core Idea

Outside derivative, bracket kept, times the derivative of the bracket.

$\dfrac{d}{dx}f(g(x)) = f'(g(x)) \times g'(x)$ . For a power of a bracket, bring the power down, lower it by $1$ and keep the bracket, then multiply by the derivative of what is inside. For a linear bracket this gives $\dfrac{d}{dx}(ax + b)^{n} = na(ax + b)^{n-1}$ . Write $k(ax + b)^{n}$ with its constant multiple in front: the answer is $kn{}a(ax + b)^{n-1}$ .

Think About It

Why is there an extra factor from the inside of the bracket? Check the rule on

(2x + 5)^{2}

by expanding it first.

Show the full workingHide the working

\displaystyle \left(2x + 5\right)^{2} = 4x^{2} + 20x + 25

Expand the bracket so the power rule can be used on each term.

\displaystyle \dfrac{d}{dx}\left(4x^{2} + 20x + 25\right) = 8x + 20 = 4\left(2x + 5\right)

Power rule term by term, then take out the common factor

4

The chain rule gives

2 \times 2\left(2x + 5\right)^{1} = 4\left(2x + 5\right)

. The first

2

is the power and the second

2

is the derivative of the bracket

2x + 5

Answer

Expanding and the chain rule both give

4(2x + 5)

, so the factor

2

from the inside of the bracket really is needed.

Example 3.1

A power of a linear bracket

Differentiate $y = \left(2 x + 5\right)^{4}$ with respect to $x$ .

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{4\left(2 x + 5\right)^{3}} \times \textcolor{#010103}{2}

Outside: bring the power down and multiply it into the coefficient, which gives

4

, and lower the power to

3

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{8 \left(2 x + 5\right)^{3}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = 8 \left(2 x + 5\right)^{3}

Do not forget the derivative of the bracket. The derivative of

(3x - 1)^{5}

15(3x - 1)^{4}

, not

5(3x - 1)^{4}

: the missing factor is the

3

from inside. Do not change the bracket either: the new power goes on the whole bracket, so

(3x - 1)^{5}

never becomes

5(3x)^{4}

Example 3.2

A negative index: the reciprocal of a bracket

Find $\dfrac{dy}{dx}$ when $y = \frac{3}{\left(4 x - 1\right)^{2}}$ .

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{-6\left(4 x - 1\right)^{-3}} \times \textcolor{#010103}{4}

Outside: bring the power down and multiply it into the coefficient, which gives

-6

, and lower the power to

-3

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{- \frac{24}{\left(4 x - 1\right)^{3}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = - \frac{24}{\left(4 x - 1\right)^{3}}

Your Turn 3.2

Find $\dfrac{dy}{dx}$ when $y = \frac{5}{\left(2 - 3 x\right)^{3}}$ .

Show full solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{-15\left(2 - 3 x\right)^{-4}} \times \textcolor{#010103}{\left(-3\right)}

Outside: bring the power down and multiply it into the coefficient, which gives

-15

, and lower the power to

-4

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{\frac{45}{\left(2 - 3 x\right)^{4}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = \frac{45}{\left(2 - 3 x\right)^{4}}

Example 3.3

The square root of a linear bracket

Differentiate $y = \sqrt{3 x + 7}$ with respect to $x$ , for values of $x$ where the root exists.

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{\frac{1}{2}\left(3 x + 7\right)^{- \frac{1}{2}}} \times \textcolor{#010103}{3}

Outside: bring the power down and multiply it into the coefficient, which gives

\frac{1}{2}

, and lower the power to

- \frac{1}{2}

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{\frac{3}{2 \sqrt{3 x + 7}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = \frac{3}{2 \sqrt{3 x + 7}}

Your Turn 3.3

Differentiate $y = \sqrt{8 - 5 x}$ with respect to $x$ , for values of $x$ where the root exists.

Show full solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{\frac{1}{2}\left(8 - 5 x\right)^{- \frac{1}{2}}} \times \textcolor{#010103}{\left(-5\right)}

Outside: bring the power down and multiply it into the coefficient, which gives

\frac{1}{2}

, and lower the power to

- \frac{1}{2}

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{- \frac{5}{2 \sqrt{8 - 5 x}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = - \frac{5}{2 \sqrt{8 - 5 x}}

The bracket does not have to be linear. For $(x^{2} - 3x)^{6}$ the derivative of the bracket is $2x - 3$ , and it multiplies the front, so the result is a product. Taking out the common factor leaves a tidy answer.

Example 3.4

A quadratic bracket, answer factorised

Differentiate $y = \left(x^{2} - 3 x\right)^{6}$ with respect to $x$ . Give your answer in factorised form.

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{6\left(x^{2} - 3 x\right)^{5}} \times \textcolor{#010103}{\left(2 x - 3\right)}

Outside: bring the power down and multiply it into the coefficient, which gives

6

, and lower the power to

5

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{6 x^{5} \left(x - 3\right)^{5} \left(2 x - 3\right)}

Multiply the numbers together.

Answer

\frac{dy}{dx} = 6 x^{5} \left(x - 3\right)^{5} \left(2 x - 3\right)

Your Turn 3.4

Differentiate $y = \left(2 x^{2} + x\right)^{5}$ with respect to $x$ . Give your answer in factorised form.

Show full solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{5\left(2 x^{2} + x\right)^{4}} \times \textcolor{#010103}{\left(4 x + 1\right)}

Outside: bring the power down and multiply it into the coefficient, which gives

5

, and lower the power to

4

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{5 x^{4} \left(2 x + 1\right)^{4} \left(4 x + 1\right)}

Multiply the numbers together.

Answer

\frac{dy}{dx} = 5 x^{4} \left(2 x + 1\right)^{4} \left(4 x + 1\right)

Example 3.5

The reciprocal square root of a quadratic

Find $\dfrac{dy}{dx}$ when $y = \frac{1}{\sqrt{x^{2} + 4}}$ .

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{- \frac{1}{2}\left(x^{2} + 4\right)^{- \frac{3}{2}}} \times \textcolor{#010103}{2 x}

Outside: bring the power down and multiply it into the coefficient, which gives

- \frac{1}{2}

, and lower the power to

- \frac{3}{2}

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{- \frac{x}{\left(x^{2} + 4\right)^{\frac{3}{2}}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = - \frac{x}{\left(x^{2} + 4\right)^{\frac{3}{2}}}

Your Turn 3.5

Find $\dfrac{dy}{dx}$ when $y = \frac{4}{\sqrt{2 x^{2} - 1}}$ .

Show full solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{-2\left(2 x^{2} - 1\right)^{- \frac{3}{2}}} \times \textcolor{#010103}{4 x}

Outside: bring the power down and multiply it into the coefficient, which gives

-2

, and lower the power to

- \frac{3}{2}

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{- \frac{8 x}{\left(2 x^{2} - 1\right)^{\frac{3}{2}}}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = - \frac{8 x}{\left(2 x^{2} - 1\right)^{\frac{3}{2}}}

SGSchoolKaki Tip

Before using the chain rule, ask: is this a power of something that is not just $x$ ? If so, the something is the bracket. A root $\sqrt{g}$ is the bracket $g$ to the power $\dfrac{1}{2}$ , and $\dfrac{1}{g}$ is $g^{-1}$ .

Example 3.6

A constant multiple of a bracket power

Find $\dfrac{dy}{dx}$ when $y = 2 \left(4 - x\right)^{5}$ .

Show solutionHide solution

Colour keyinside function or factor brought inderivative of a term or factorfinal derivative

\displaystyle \frac{dy}{dx}\;= \textcolor{#010105}{10\left(4 - x\right)^{4}} \times \textcolor{#010103}{\left(-1\right)}

Outside: bring the power down and multiply it into the coefficient, which gives

10

, and lower the power to

4

, keeping the bracket. Inside: multiply by the derivative of the bracket.

\displaystyle = \textcolor{#010106}{- 10 \left(4 - x\right)^{4}}

Multiply the numbers together.

Answer

\frac{dy}{dx} = - 10 \left(4 - x\right)^{4}

Links To

Derivatives of sine, cosine, tangent, exponential and logarithmic functions use these same rules, especially the chain rule, and are the subject of Chapter 13: Differentiation of trigonometric, exponential and logarithmic functions

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 11 of 17