Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Gradients and the power rule

Pure content (sections A–J) · Subtopic 22 of 66

Gradients and the Power Rule

What the gradient of a curve means, how chords close in on the tangent, and the power rule $\dfrac{d}{dx}x^n=nx^{n-1}$ that makes differentiating fast, including roots, fractions and second derivatives.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Explain the gradient of a curve at a point as the limit of the gradients of chords.
Differentiate $ax^n$ for any rational $n$ , and sums and differences of such terms.
Rewrite roots, fractions and products as powers of $x$ before differentiating.
Find gradients at points, points with a given gradient, and second derivatives.

Pause and think

A straight line has the same gradient everywhere. The curve $y=x^2$ gets steeper as $x$ increases. How could you measure "how steep" it is at exactly $x=1$ ?

Section 2 of 14

Core idea

Start hereThe speedometer problem

A car's distance from home is $s=t^2$ metres after $t$ seconds. Its average speed over the first $3$ seconds is $\frac93=3$ m/s. But what does the speedometer read at the exact instant $t=1$ ? Speed is distance ÷ time, and at an instant no time passes: $\frac00$ . For 2000 years that looked impossible. Newton and Leibniz cracked it by taking smaller and smaller time intervals and seeing what the average speed approaches. That idea is the derivative.

The big idea

The derivative $\dfrac{dy}{dx}$ is the gradient of a curve at a point: the limit of chord gradients as the chord shrinks. For powers of $x$ , $\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}$ .

1. The problem: a curve has no single gradient

A straight line has the same steepness everywhere. A curve like $y=x^2$ gets steeper as you move right. "The gradient 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.

Key takeawayThe central problem A gradient needs two points (

\frac{\text{rise}}{\text{run}}

). A tangent is defined by one. Solution: use a second point very close by, then let it slide in.

2. Chords closing in on the tangent

Chords approaching the tangent at (1, 1)

gap h = 2.00

The chord from

(1,1)

(1+h,(1+h)^2)

has gradient

\dfrac{(1+h)^2-1}{h}=2+h

. As

h

shrinks to

0

it becomes

2

: the tangent at

x=1

has gradient

2

Gradient of

y=x^2

x=1

$P(1,1),\quad Q(1+h,\,(1+h)^2)$
$Q$ is a nearby point, $h$ to the right ( $h\neq0$ ).
$m_{PQ}=\frac{(1+h)^2-1}{(1+h)-1}$
Chord gradient: rise over run.
$=\frac{1+2h+h^2-1}{h}$
Expand $(1+h)^2$ .
$=\frac{2h+h^2}{h}=2+h$
Simplify; divide by $h$ (allowed since $h\neq0$ ).
$\lim_{h\to0}(2+h)=2$
As $Q$ slides to $P$ , $h$ shrinks to $0$ .

The gradient of

y=x^2

x=1

is exactly

2

. (So the car's speed at

t=1

2

m/s.)

Do the same at a general point $x$ to get a formula for the gradient everywhere. This is called differentiation from first principles.

Definition of the derivative

f'(x)=\frac{dy}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}

$h$: a small step in $x$ .
$\lim_{h\to0}$: "the value approached as $h$ gets closer and closer to $0$ ".
$\frac{dy}{dx}$ , $f'(x)$: two notations for the derivative (gradient function).

First principles for

y=x^3

$\frac{(x+h)^3-x^3}{h}$
Chord gradient from $x$ to $x+h$ .
$=\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}$
Expand $(x+h)^3$ (binomial).
$=\frac{3x^2h+3xh^2+h^3}h=3x^2+3xh+h^2$
Cancel $x^3$ ; divide by $h$ .
$\to3x^2\text{ as }h\to0$
Terms with $h$ vanish.

\frac{d}{dx}x^3=3x^2

A curve and its gradient function

Left of

0

the curve goes downhill, and the gradient function is negative. At the bottom it is flat, gradient

0

. To the right it is uphill and the gradient is positive and growing.

3. The power rule

$x^2\to2x$ , $x^3\to3x^2$ ; in general (using the binomial expansion of $(x+h)^n$ , whose second term is $nx^{n-1}h$ ):

Power rule

\frac{d}{dx}\left(x^n\right)=nx^{n-1}\qquad\frac{d}{dx}(ax^n)=anx^{n-1}\qquad\frac{d}{dx}(c)=0

$n$: any real power: positive, negative or fractional.
$c$: a constant; its graph is flat, so gradient 0.

Differentiate a sum term by term. Before differentiating, rewrite roots and fractions as powers: $\sqrt x=x^{1/2}$ , $\frac1{x^2}=x^{-2}$ .

Worked example $y=4x^3-\dfrac2{x^2}+3\sqrt x-7$

Find $\dfrac{dy}{dx}$ .

Goal: The gradient function.
Strategy: Rewrite every term as a power of $x$ , then apply the power rule to each.

$y=4x^3-2x^{-2}+3x^{1/2}-7$
Rewrite.
$\frac{dy}{dx}=12x^2-2(-2)x^{-3}+3\cdot\tfrac12x^{-1/2}-0$
Bring the power down; reduce it by one.
$=12x^2+\frac4{x^3}+\frac3{2\sqrt x}$
Tidy.

Answer

12x^2+\dfrac4{x^3}+\dfrac3{2\sqrt x}

✓ CheckThe constant

-7

disappeared, as it should (shifting a graph up doesn't change its steepness).

Worked exampleGradient at a point, and where the gradient is given

For $y=x^3-6x^2+5$ : find the gradient at $x=1$ , the points where the gradient is $15$ , and $\frac{d^2y}{dx^2}$ .

Goal: Three results.
Strategy: Differentiate; substitute or solve.

$\frac{dy}{dx}=3x^2-12x$
Power rule.
$x=1:\ 3-12=-9$
Substitute.
$3x^2-12x=15\Rightarrow x^2-4x-5=0\Rightarrow(x-5)(x+1)=0$
Set equal and solve.
$\frac{d^2y}{dx^2}=6x-12$
Differentiate again: the second derivative, the rate of change of the gradient.

Answer

-9

;

x=5

x=-1

;

6x-12

✓ CheckAt

x=5

75-60=15

✓.

Common misconceptionProducts and quotients

\frac{d}{dx}\left[(x^2+1)(x-3)\right]

is not

2x\times1

. Expand first (or use the product rule later in the course).

Pause and think

Find $f'(4)$ for $f(x)=\dfrac8{\sqrt x}$ .

Summary

Derivative = limit of chord gradients.
$\frac{d}{dx}x^n=nx^{n-1}$ ; constants vanish.
Rewrite as powers first.
$\frac{d^2y}{dx^2}$ : differentiate twice.

Derivative: Gradient function.
Tangent: Line touching a curve, with its gradient.
Limit: The value approached.
First principles: Deriving the gradient from chords.

HistoryTwo inventors, one bitter feud. Isaac Newton developed calculus in the 1660s (he called it the method of fluxions) and Gottfried Leibniz independently in the 1670s. Their supporters argued for decades over who came first. Today we credit both, and we use Leibniz's notation

\dfrac{dy}{dx}

because it is so practical.

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