1.7.5 Second derivative as rate of change of gradient

What you'll learn

How the first derivative describes the gradient of a curve.
How to calculate the second derivative by differentiating twice.
How to interpret the second derivative as the rate at which gradient changes.
What positive, negative and zero values of the second derivative tell you about a curve.

Prerequisite: the first derivative

For a curve with equation $y = f (x)$ , the gradient usually changes as you move along the curve. Unlike the gradient of a straight line, it is not necessarily constant.

The first derivative gives the gradient of the curve at a particular value of $x$ .

Definition

First derivative

The first derivative of $y$ with respect to $x$ can be written as

dydx,f′(x),ory′.\frac{dy}{dx}, \qquad f'(x), \qquad \text{or} \qquad y'.

Its value at a point is the gradient of the tangent to the curve at that point.

For example, if

y=x^3,

then

dydx=3x2.\frac{dy}{dx}=3x^2.

This tells you that the gradient depends on $x$ . At $x = 1$ , the gradient is 3, while at $x = 2$ , the gradient is 12.

Example

Finding the gradient at a point

Find the gradient of $y=2x^3-5x^2+4$ at $x = 2$ .

Differentiate the function using the power rule:
$dydx=6x2−10x.\frac{dy}{dx}=6x^2-10x.$
Substitute $x = 2$ into the derivative:
$dydx∣x=2=6(2)2−10(2).\left.\frac{dy}{dx}\right|_{x=2} =6(2)^2-10(2).$
Evaluate the expression:
$dydx∣x=2=24−20=4.\left.\frac{dy}{dx}\right|_{x=2} =24-20=4.$
The gradient of the curve at $x = 2$ is 4.

Why does the gradient itself change?

The first derivative $dydx\frac{dy}{dx}$ is normally a function of $x$ . This means that, as $x$ changes, the gradient may also change.

For the curve $y=x^2$ ,

dydx=2x.\frac{dy}{dx}=2x.

As $x$ increases, the gradient increases. For example:

at $x = - 1$ , the gradient is $- 2$ ;
at $x = 0$ , the gradient is 0;
at $x = 1$ , the gradient is 2.

The diagram shows this increasing gradient through tangents drawn at different points.

Graph of y equals x squared with tangent gradients minus 2, 0 and 2, showing that the gradient increases as x increases

The natural next question is: how quickly is the gradient changing? To answer this, you differentiate the gradient.

The second derivative

The second derivative is found by differentiating the first derivative with respect to $x$ .

Definition

Second derivative

The second derivative of $y$ with respect to $x$ is written as

d2ydx2,f′′(x),ory′′.\frac{d^2y}{dx^2}, \qquad f''(x), \qquad \text{or} \qquad y''.

It measures the rate of change of the gradient $dydx\frac{dy}{dx}$ with respect to $x$ :

d2ydx2=ddx(dydx).\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).

The notation $d2ydx2\frac{d^2y}{dx^2}$ is read as “dee two $y$ by dee $x$ squared”.

It does not mean that $dydx\frac{dy}{dx}$ has simply been squared.

Common Mistake

Do not square the first derivative

In general,

d2ydx2≠(dydx)2.\frac{d^2y}{dx^2}\neq \left(\frac{dy}{dx}\right)^2.

You must differentiate $dydx\frac{dy}{dx}$ again.

Differentiating twice

To find the second derivative:

Differentiate $y$ once to find $dydx\frac{dy}{dx}$ .
Differentiate $dydx\frac{dy}{dx}$ to find $d2ydx2\frac{d^2y}{dx^2}$ .

Example

Calculating a second derivative

Find the second derivative of

y=3x^4-5x^3+2x-7.

Differentiate $y$ once:
$dydx=12x3−15x2+2.\frac{dy}{dx}=12x^3-15x^2+2.$
Differentiate the first derivative:
$d2ydx2=36x2−30x.\frac{d^2y}{dx^2}=36x^2-30x.$
If a value of $x$ were given, you could now substitute it into $36x^2-30x$ to find the rate of change of gradient at that point.

Tip

Keep the derivative lines separate

When differentiating twice, write $y$ , $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ on separate lines. This makes it much easier to spot a missing term or power-rule error.

Interpreting the second derivative

The sign of $d2ydx2\frac{d^2y}{dx^2}$ tells you whether the gradient is increasing or decreasing as $x$ increases.

Positive second derivative

d2ydx2>0,\frac{d^2y}{dx^2}>0,

then the gradient is increasing.

The curve is often described as concave upwards: its gradient becomes more positive, or less negative, as you move from left to right.

For example, for $y=x^2$ ,

dydx=2xandd2ydx2=2.\frac{dy}{dx}=2x \quad\text{and}\quad \frac{d^2y}{dx^2}=2.

The positive second derivative confirms that the gradient increases at a constant rate.

Negative second derivative

d2ydx2<0,\frac{d^2y}{dx^2}<0,

then the gradient is decreasing.

The curve is often described as concave downwards: its gradient becomes less positive, or more negative, as you move from left to right.

Example

Interpreting a changing gradient

For the curve

y=x^3-6x^2+5x,

determine whether the gradient is increasing or decreasing at $x = 1$ and at $x = 3$ .

Differentiate twice:
$dydx=3x2−12x+5,d2ydx2=6x−12.\begin{aligned} \frac{dy}{dx}&=3x^2-12x+5,\\ \frac{d^2y}{dx^2}&=6x-12. \end{aligned}$
At $x = 1$ , evaluate the second derivative:
$d2ydx2∣x=1=6(1)−12=−6.\left.\frac{d^2y}{dx^2}\right|_{x=1} =6(1)-12=-6.$
Since this is negative, the gradient is decreasing at $x = 1$ .
At $x = 3$ , evaluate the second derivative:
$d2ydx2∣x=3=6(3)−12=6.\left.\frac{d^2y}{dx^2}\right|_{x=3} =6(3)-12=6.$
Since this is positive, the gradient is increasing at $x = 3$ .

Key Idea

The central interpretation

The first derivative tells you what the gradient is. The second derivative tells you how the gradient is changing.

When the second derivative is zero

d2ydx2=0,\frac{d^2y}{dx^2}=0,

then the gradient has an instantaneous rate of change of zero at that value of $x$ .

This does not automatically mean that the graph is a straight line or that the gradient is zero. Remember that $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ measure different things.

Example

A zero rate of change of gradient

Consider

y=x^3.

Differentiate twice:
$dydx=3x2,d2ydx2=6x.\begin{aligned} \frac{dy}{dx}&=3x^2,\\ \frac{d^2y}{dx^2}&=6x. \end{aligned}$
At $x = 0$ ,
$d2ydx2∣x=0=0.\left.\frac{d^2y}{dx^2}\right|_{x=0}=0.$
On the left of zero, $6 x < 0$ , so the gradient is decreasing. On the right of zero, $6 x > 0$ , so the gradient is increasing. The way the curve bends changes at $x = 0$ .

A point where the direction of curvature changes is called a point of inflection. A zero second derivative identifies a possible point of inflection, but you must check that the sign of $d2ydx2\frac{d^2y}{dx^2}$ actually changes.

Common Mistake

Zero is not enough

The condition $d2ydx2=0\frac{d^2y}{dx^2}=0$ alone does not prove that a point is a point of inflection. You need evidence that the curvature changes direction, usually by checking the sign on either side.

The size of the second derivative

The magnitude of $d2ydx2\frac{d^2y}{dx^2}$ describes how rapidly the gradient changes.

For example:

a second derivative of 20 means the gradient increases rapidly;
a second derivative of $- 20$ means the gradient decreases rapidly;
a second derivative close to zero means the gradient changes slowly near that point.

Always consider both the sign and the magnitude.

Connection with motion

When displacement $s$ is a function of time $t$ ,

dsdt=v\frac{ds}{dt}=v

is velocity, which is the gradient of a displacement–time graph.

Differentiating again gives

d2sdt2=dvdt=a,\frac{d^2s}{dt^2} = \frac{dv}{dt} = a,

where $a$ is acceleration. Acceleration is therefore the rate of change of velocity, just as the second derivative of a general function is the rate of change of its gradient.

Example

Interpreting the second derivative in motion

A particle has displacement

s=t^3-6t^2+9t

metres after $t$ seconds. Find its acceleration when $t = 3$ .

Differentiate displacement to obtain velocity:
$v=dsdt=3t2−12t+9.v=\frac{ds}{dt}=3t^2-12t+9.$
Differentiate velocity to obtain acceleration:
$a=dvdt=d2sdt2=6t−12.a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=6t-12.$
Substitute $t = 3$ :
$s−2.a=6(3)-12=6\text{ m s}^{-2}.$
At this instant, the velocity is increasing at a rate of 6 m s⁻².

Exam technique

In the exam

Write both derivative stages clearly, even if you could calculate the second derivative mentally.
If asked for an interpretation, state whether the gradient is increasing or decreasing, not merely that the second derivative is positive or negative.
When $d2ydx2=0\frac{d^2y}{dx^2}=0$ , check values on either side before claiming that there is a point of inflection.
In a motion problem, include the correct units: velocity in m s⁻¹ and acceleration in m s⁻².

Self review

Check yourself

What is the difference between $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ ?
Find $d2ydx2\frac{d^2y}{dx^2}$ for $y=2x^5-3x^3+7x$ .
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ at a point, what does this tell you about the gradient there?

What you'll learn

How the first derivative describes the gradient of a curve.
How to calculate the second derivative by differentiating twice.
How to interpret the second derivative as the rate at which gradient changes.
What positive, negative and zero values of the second derivative tell you about a curve.

Prerequisite: the first derivative

For a curve with equation $y = f (x)$ , the gradient usually changes as you move along the curve. Unlike the gradient of a straight line, it is not necessarily constant.

The first derivative gives the gradient of the curve at a particular value of $x$ .

Definition

First derivative

The first derivative of $y$ with respect to $x$ can be written as

dydx,f′(x),ory′.\frac{dy}{dx}, \qquad f'(x), \qquad \text{or} \qquad y'.

Its value at a point is the gradient of the tangent to the curve at that point.

For example, if

y=x^3,

then

dydx=3x2.\frac{dy}{dx}=3x^2.

This tells you that the gradient depends on $x$ . At $x = 1$ , the gradient is 3, while at $x = 2$ , the gradient is 12.

Example

Finding the gradient at a point

Find the gradient of $y=2x^3-5x^2+4$ at $x = 2$ .

Differentiate the function using the power rule:
$dydx=6x2−10x.\frac{dy}{dx}=6x^2-10x.$
Substitute $x = 2$ into the derivative:
$dydx∣x=2=6(2)2−10(2).\left.\frac{dy}{dx}\right|_{x=2} =6(2)^2-10(2).$
Evaluate the expression:
$dydx∣x=2=24−20=4.\left.\frac{dy}{dx}\right|_{x=2} =24-20=4.$
The gradient of the curve at $x = 2$ is 4.

Why does the gradient itself change?

The first derivative $dydx\frac{dy}{dx}$ is normally a function of $x$ . This means that, as $x$ changes, the gradient may also change.

For the curve $y=x^2$ ,

dydx=2x.\frac{dy}{dx}=2x.

As $x$ increases, the gradient increases. For example:

at $x = - 1$ , the gradient is $- 2$ ;
at $x = 0$ , the gradient is 0;
at $x = 1$ , the gradient is 2.

The diagram shows this increasing gradient through tangents drawn at different points.

Graph of y equals x squared with tangent gradients minus 2, 0 and 2, showing that the gradient increases as x increases

The natural next question is: how quickly is the gradient changing? To answer this, you differentiate the gradient.

The second derivative

The second derivative is found by differentiating the first derivative with respect to $x$ .

Definition

Second derivative

The second derivative of $y$ with respect to $x$ is written as

d2ydx2,f′′(x),ory′′.\frac{d^2y}{dx^2}, \qquad f''(x), \qquad \text{or} \qquad y''.

It measures the rate of change of the gradient $dydx\frac{dy}{dx}$ with respect to $x$ :

d2ydx2=ddx(dydx).\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).

The notation $d2ydx2\frac{d^2y}{dx^2}$ is read as “dee two $y$ by dee $x$ squared”.

It does not mean that $dydx\frac{dy}{dx}$ has simply been squared.

Common Mistake

Do not square the first derivative

In general,

d2ydx2≠(dydx)2.\frac{d^2y}{dx^2}\neq \left(\frac{dy}{dx}\right)^2.

You must differentiate $dydx\frac{dy}{dx}$ again.

Differentiating twice

To find the second derivative:

Differentiate $y$ once to find $dydx\frac{dy}{dx}$ .
Differentiate $dydx\frac{dy}{dx}$ to find $d2ydx2\frac{d^2y}{dx^2}$ .

Example

Calculating a second derivative

Find the second derivative of

y=3x^4-5x^3+2x-7.

Differentiate $y$ once:
$dydx=12x3−15x2+2.\frac{dy}{dx}=12x^3-15x^2+2.$
Differentiate the first derivative:
$d2ydx2=36x2−30x.\frac{d^2y}{dx^2}=36x^2-30x.$
If a value of $x$ were given, you could now substitute it into $36x^2-30x$ to find the rate of change of gradient at that point.

Tip

Keep the derivative lines separate

When differentiating twice, write $y$ , $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ on separate lines. This makes it much easier to spot a missing term or power-rule error.

Interpreting the second derivative

The sign of $d2ydx2\frac{d^2y}{dx^2}$ tells you whether the gradient is increasing or decreasing as $x$ increases.

Positive second derivative

d2ydx2>0,\frac{d^2y}{dx^2}>0,

then the gradient is increasing.

The curve is often described as concave upwards: its gradient becomes more positive, or less negative, as you move from left to right.

For example, for $y=x^2$ ,

dydx=2xandd2ydx2=2.\frac{dy}{dx}=2x \quad\text{and}\quad \frac{d^2y}{dx^2}=2.

The positive second derivative confirms that the gradient increases at a constant rate.

Negative second derivative

d2ydx2<0,\frac{d^2y}{dx^2}<0,

then the gradient is decreasing.

The curve is often described as concave downwards: its gradient becomes less positive, or more negative, as you move from left to right.

Example

Interpreting a changing gradient

For the curve

y=x^3-6x^2+5x,

determine whether the gradient is increasing or decreasing at $x = 1$ and at $x = 3$ .

Differentiate twice:
$dydx=3x2−12x+5,d2ydx2=6x−12.\begin{aligned} \frac{dy}{dx}&=3x^2-12x+5,\\ \frac{d^2y}{dx^2}&=6x-12. \end{aligned}$
At $x = 1$ , evaluate the second derivative:
$d2ydx2∣x=1=6(1)−12=−6.\left.\frac{d^2y}{dx^2}\right|_{x=1} =6(1)-12=-6.$
Since this is negative, the gradient is decreasing at $x = 1$ .
At $x = 3$ , evaluate the second derivative:
$d2ydx2∣x=3=6(3)−12=6.\left.\frac{d^2y}{dx^2}\right|_{x=3} =6(3)-12=6.$
Since this is positive, the gradient is increasing at $x = 3$ .

Key Idea

The central interpretation

The first derivative tells you what the gradient is. The second derivative tells you how the gradient is changing.

When the second derivative is zero

d2ydx2=0,\frac{d^2y}{dx^2}=0,

then the gradient has an instantaneous rate of change of zero at that value of $x$ .

This does not automatically mean that the graph is a straight line or that the gradient is zero. Remember that $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ measure different things.

Example

A zero rate of change of gradient

Consider

y=x^3.

Differentiate twice:
$dydx=3x2,d2ydx2=6x.\begin{aligned} \frac{dy}{dx}&=3x^2,\\ \frac{d^2y}{dx^2}&=6x. \end{aligned}$
At $x = 0$ ,
$d2ydx2∣x=0=0.\left.\frac{d^2y}{dx^2}\right|_{x=0}=0.$
On the left of zero, $6 x < 0$ , so the gradient is decreasing. On the right of zero, $6 x > 0$ , so the gradient is increasing. The way the curve bends changes at $x = 0$ .

Common Mistake

Zero is not enough

The size of the second derivative

The magnitude of $d2ydx2\frac{d^2y}{dx^2}$ describes how rapidly the gradient changes.

For example:

a second derivative of 20 means the gradient increases rapidly;
a second derivative of $- 20$ means the gradient decreases rapidly;
a second derivative close to zero means the gradient changes slowly near that point.

Always consider both the sign and the magnitude.

Connection with motion

When displacement $s$ is a function of time $t$ ,

dsdt=v\frac{ds}{dt}=v

is velocity, which is the gradient of a displacement–time graph.

Differentiating again gives

d2sdt2=dvdt=a,\frac{d^2s}{dt^2} = \frac{dv}{dt} = a,

where $a$ is acceleration. Acceleration is therefore the rate of change of velocity, just as the second derivative of a general function is the rate of change of its gradient.

Example

Interpreting the second derivative in motion

A particle has displacement

s=t^3-6t^2+9t

metres after $t$ seconds. Find its acceleration when $t = 3$ .

Differentiate displacement to obtain velocity:
$v=dsdt=3t2−12t+9.v=\frac{ds}{dt}=3t^2-12t+9.$
Differentiate velocity to obtain acceleration:
$a=dvdt=d2sdt2=6t−12.a=\frac{dv}{dt}=\frac{d^2s}{dt^2}=6t-12.$
Substitute $t = 3$ :
$s−2.a=6(3)-12=6\text{ m s}^{-2}.$
At this instant, the velocity is increasing at a rate of 6 m s⁻².

Exam technique

In the exam

Write both derivative stages clearly, even if you could calculate the second derivative mentally.
If asked for an interpretation, state whether the gradient is increasing or decreasing, not merely that the second derivative is positive or negative.
When $d2ydx2=0\frac{d^2y}{dx^2}=0$ , check values on either side before claiming that there is a point of inflection.
In a motion problem, include the correct units: velocity in m s⁻¹ and acceleration in m s⁻².

Self review

Check yourself

What is the difference between $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ ?
Find $d2ydx2\frac{d^2y}{dx^2}$ for $y=2x^5-3x^3+7x$ .
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ at a point, what does this tell you about the gradient there?

What you'll learn

Prerequisite: the first derivative

Why does the gradient itself change?

The second derivative

Differentiating twice

Interpreting the second derivative

Positive second derivative

Negative second derivative

When the second derivative is zero

The size of the second derivative

Connection with motion

Review 1.7.5 Second derivative as rate of change of gradient by teaching Genie

1.7.5 Second derivative as rate of change of gradient

What you'll learn

Prerequisite: the first derivative

Why does the gradient itself change?

The second derivative

Differentiating twice

Interpreting the second derivative

Positive second derivative

Negative second derivative

When the second derivative is zero

The size of the second derivative

Connection with motion

Review 1.7.5 Second derivative as rate of change of gradient by teaching Genie

1.7.5 Second derivative as rate of change of gradient

What you'll learn

Prerequisite: the first derivative

Why does the gradient itself change?

The second derivative

Differentiating twice

Interpreting the second derivative

Positive second derivative

Negative second derivative

When the second derivative is zero

The size of the second derivative

Connection with motion

Review 1.7.5 Second derivative as rate of change of gradient by teaching Genie

1.7.5 Second derivative as rate of change of gradient Revision Guide

1.7.5 Second derivative as rate of change of gradient

What you'll learn

Prerequisite: the first derivative

Why does the gradient itself change?

The second derivative

Differentiating twice

Interpreting the second derivative

Positive second derivative

Negative second derivative

When the second derivative is zero

The size of the second derivative

Connection with motion

Review 1.7.5 Second derivative as rate of change of gradient by teaching Genie

1.7.5 Second derivative as rate of change of gradient Revision Guide