LearnYMA01WMA11 · Pure Mathematics 1DifferentiationGradients, rates of change and second derivatives

Differentiation · Subtopic 3 of 4

Gradients, Rates of Change and Second Derivatives

Substitute an $x$ -value into $\dfrac{dy}{dx}$ to get the gradient there; set $\dfrac{dy}{dx}$ equal to a number to find where the curve has that gradient. Differentiating again gives the second derivative $\dfrac{d^2y}{dx^2}$ , the rate at which the gradient itself changes.

Edexcel YMA01 WMA11≈ 30 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Find the gradient of a curve at a given point.
Find the points where the gradient has a given value.
Interpret $\dfrac{dy}{dx}$ as a rate of change.
Find and evaluate second derivatives.

Pause and think

A ball's height is $h$ metres after $t$ seconds. What would $\dfrac{dh}{dt}$ tell you? And $\dfrac{d^2h}{dt^2}$ ?

Section 2 of 14

Core idea

Start hereThe top of the hill

At the very top of a hill the ground is momentarily flat. Find where a curve's gradient is zero and you've found its peaks and valleys: the key to every "maximum profit" or "least material" problem.

The big idea

$\frac{dy}{dx}>0$ increasing, $<0$ decreasing, $=0$ stationary. Classify with $\frac{d^2y}{dx^2}$ : negative = maximum, positive = minimum.

1. Stationary points

y = x³ − 3x and its gradient function

The gradient function is zero at

x=\pm1

, exactly where the curve has its turning points. Between them it is negative, and the curve goes downhill.

Finding and classifying

Drawing diagram…

$\frac{d^2y}{dx^2}$ tells whether the gradient is falling (max) or rising (min).

Worked example $y=x^3-3x$

Stationary points, their nature, and where $y$ is decreasing.

Goal: Points; interval.
Strategy: Flowchart.

$\frac{dy}{dx}=3x^2-3=3(x-1)(x+1)$
Differentiate and factorise.
$x=\pm1:\ (-1,2),\ (1,-2)$
$y(-1)=-1+3$ , $y(1)=1-3$ .
$\frac{d^2y}{dx^2}=6x:\ -6\ (\max),\ +6\ (\min)$
Signs.
$3(x-1)(x+1)<0\iff-1<x<1$
Decreasing where gradient negative.

AnswerMax

(-1,2)

, min

(1,-2)

; decreasing for

-1<x<1

✓ CheckPositive cubic: max comes before min ✓.

Worked exampleAlways increasing

Show $f(x)=x^3+3x$ is increasing for all $x$ .

Goal: Proof.
Strategy: Show $f'(x)>0$ always.

$f'(x)=3x^2+3$
Differentiate.
$x^2\ge0\Rightarrow3x^2+3\ge3>0$
Squares are non-negative.

Answer

f'(x)>0

for all

x

, so

f

is increasing.

✓ CheckNo stationary points:

3x^2+3=0

has no real solution ✓.

Edge case $\frac{d^2y}{dx^2}=0$ Inconclusive:

y=x^4

(min) and

y=x^3

(inflection) both give

0

at the origin.

Pause and think

Stationary point of $y=x^2-6x$ ?

Summary

Gradient zero at stationary points.
Second derivative classifies.

Stationary point: Where the gradient is zero.

HistoryFluxions. Newton called derivatives "fluxions" and thought of quantities as flowing in time. He developed the idea during 1665–1666, when Cambridge closed because of the plague and he worked at home in Lincolnshire.

Fun factJerk, snap, crackle, pop. The derivative of acceleration is called jerk. Physicists have jokingly named the next derivatives snap, crackle and pop. Lift and rollercoaster designers keep jerk low so rides feel smooth.

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