4.03 Using the second derivative

The second derivative

Taking the second derivative means differentiating a function twice or finding the derivative of the first derivative. In fact, if the function allows, we can take the third, fourth, fifth and onwards derivatives.

Let's have a look at what this means for a simple function.

If we start with a function like:

$y$y

$=$=

$x^5$x5

Then the first derivative is:

$y'$y′

$=$=

$5x^4$5x4

We can take the derivative again and call this the second derivative, and use the notation $y''$y′′:

$y''$y′′

$=$=

$20x^3$20x3

 

We can keep going and going if we wanted to, adding a prime to the notation each time.

$y'''$y′′′

$=$=

$60x^2$60x2

 

In this course, it is only useful for us to consider up to the second derivative. We will use it to determine the nature and behaviour of functions and their important points.

As with functions and the first derivative, there are multiple notations that can be used for the second derivative. You should always use the same notation that has been provided. The table below summarises the different possibilities:

Practice questions

Question 1

The second derivative and concavity

Just as the first derivative gives the rate of change of a function, the second derivative is used to determine if the gradient at any point is either increasing, decreasing or stationary (not increasing or decreasing). The following table summarises the three possibilities:

If$f''(x)>0$f′′(x)>0	$f'(x)$f′(x) is increasing
If$f''(x)<0$f′′(x)<0	$f'(x)$f′(x) is decreasing
If$f''(x)=0$f′′(x)=0	$f'(x)$f′(x) is stationary

Concavity

Using the three conditions for the second derivative shown in the table above we can classify the concavity of a function as concave up, concave down or zero concavity. We will look at each possible combination of first and second derivatives in more detail below.

For $f''(x)>0$f′′(x)>0 and $f'(x)>0$f′(x)>0, the curve is considered concave upwards. In the diagram below, notice that the tangents have a positive gradient and are increasing in steepness from left to right. We say that the curve is increasing at an increasing rate.

For $f''(x)>0$f′′(x)>0 and $f'(x)<0$f′(x)<0, the curve is considered concave upwards as well. In the diagram below, notice that the tangents have a negative gradient and are decreasing in steepness from left to right. We say that the curve is decreasing at a decreasing rate.

For $f''(x)<0$f′′(x)<0 and $f'(x)>0$f′(x)>0, the curve is considered concave downwards. In the diagram below, notice that the tangents have a positive gradient and are decreasing in steepness from left to right. We say the curve is increasing at a decreasing rate.

For $f''(x)<0$f′′(x)<0 and $f'(x)<0$f′(x)<0, the curve is considered concave downwards. In the diagram below, notice that the tangents have a negative gradient and are increasing in steepness from left to right. We say that the curve is decreasing at an increasing rate.

For $f''(x)=0$f′′(x)=0, the curve is considered neither concave up or down at this point, that is, it has a concavity of zero. The magnitude of the gradient $f'(x)$f′(x) at this point is neither increasing or decreasing. As we can see below when $f''(x)=0$f′′(x)=0, there is no change in the sign of the gradient, but there is a change in the sign of the second derivative, hence we have a change in concavity. This point is called a point of inflection, similar to a stationary or horizontal point of inflection but without the zero gradient.

It's important to note that a point of infection isn't just when $f''\left(x\right)=0$f′′(x)=0. We need to confirm that the concavity changes on either side too. This is different from a stationary point since where we only care that $f'\left(x\right)=0$f′(x)=0.

To summarise:

Worked examples

Example 1

Find where the function $f(x)=2x^3-9x^2-108x+2$f(x)=2x3−9x2−108x+2 is concave up and concave down.

Think: Concavity is determined according to the sign of the second derivative.

Do: Find the second derivative:

$f'(x)$f′(x)	$=$=	$6x^2-18x-108$6x2−18x−108
$f''(x)$f′′(x)	$=$=	$12x-18$12x−18

 

To find where the curve is concave down solve $f''(x)<0$f′′(x)<0:

$12x-18$12x−18	$<$<	$0$0
$12x$12x	$<$<	$18$18
$x$x	$<$<	$\frac{3}{2}$32​

 

To find where the curve is concave up we solve $f''(x)>0$f′′(x)>0:

$12x-18$12x−18	$>$>	$0$0
$12x$12x	$>$>	$18$18
$x$x	$>$>	$\frac{3}{2}$32​

 

So the curve is concave down for $x<\frac{3}{2}$x<32​ and concave up for $x>\frac{3}{2}$x>32​.

Classifying minimum, maximum and inflection points

Using the first and second derivative together provides a more efficient method of determining whether a stationary point is a minimum or maximum turning point or horizontal point of inflection.

For $f'(x)=0$f′(x)=0 and $f''(x)>0$f′′(x)>0 we will always have a minimum turning point. The first derivative tells us that it is a stationary point and the second derivative tells us that the curve is concave up at this point.

For $f'(x)=0$f′(x)=0 and $f''(x)<0$f′′(x)<0 we will always have a maximum turning point. The first derivative tells us that it is a stationary point and the second derivative tells us that the curve is concave down at this point.

For $f'(x)=0$f′(x)=0 and $f''(x)=0$f′′(x)=0 we will have a stationary or horizontal point of inflection if there is a change in concavity at the point. It is important to always check the concavity if this situation arises as will be shown in the example below.

For the case where $f'(x)\ne0$f′(x)≠0 and $f''(x)=0$f′′(x)=0 or $f''(x)$f′′(x) is undefined, we will have a point of inflection if there is a change in concavity at that point. Again, It is important to always check the concavity if this situation arises.

Worked examples

Example 2

Find and classify all stationary points on the curve $f(x)=x^4-2x^2+3$f(x)=x4−2x2+3 .

Think: To find the stationary points, we want to determine the values of $x$x such that $f'\left(x\right)=0$f′(x)=0. To classify them, we can refer to the second derivative.

Do: The first derivative is:

$f'(x)$f′(x)	$=$=	$4x^3-4x$4x3−4x
$f'(x)$f′(x)	$=$=	$4x(x^2-1)$4x(x2−1)
$f'(x)$f′(x)	$=$=	$4x(x+1)(x-1)$4x(x+1)(x−1)

 

Make $f'(x)=0$f′(x)=0 to find stationary points:

$f'(x)$f′(x)	$=$=	$0$0
$4x(x+1)(x-1)$4x(x+1)(x−1)	$=$=	$0$0
$x$x	$=$=	$-1,0,1$−1,0,1

 

We can see that there are stationary points at $x=-1,0,1$x=−1,0,1. We now need to calculate the second derivative for each of these values to determine the nature of the stationary points.

$x$x	$-1$−1	$0$0	$1$1
$f''(x)$f′′(x)	$8$8	$-4$−4	$8$8
Concavity	positive	negative	positive
Meaning	concave up	concave down	concave up
Turning point classification	minimum	maximum	minimum

Graph verification. Let's check if our findings are confirmed with a graph of the function.

All that's left now is to fully state the points and their classification:

At $x=-1$x=−1 $f''(x)=8$f′′(x)=8, hence the curve is concave up at this point. Therefore $(-1,2)$(−1,2) is a minimum turning point.

At $x=0$x=0 $f''(x)=-4$f′′(x)=−4, hence the curve is concave down at this point. Therefore $(0,3)$(0,3) is a maximum turning point.

At $x=1$x=1 $f''(x)=8$f′′(x)=8, hence the curve is concave up at this point. Therefore $(1,2)$(1,2) is a minimum turning point.

Example 3

Find any possible points of inflection on the curve $y=x^6$y=x6.

Think: We first want to go straight to the second derivative and determine the values of $x$x such that $y''=0$y′′=0. Then we want to determine if these points have a change of concavity.

Do: Taking the first and second derivatives:

$y'$y′	$=$=	$6x^5$6x5
$y''$y′′	$=$=	$30x^4$30x4

 

As possible points of inflection occur where $y''=0$y′′=0 we can see there is a possible point of inflection at $x=0$x=0.

We will now check the second derivative either side of this point to check concavity and determine its nature:

$x$x	$-1$−1	$0$0	$1$1
$f''(x)$f′′(x)	$30$30	$0$0	$30$30
Concavity	positive	zero	negative
Meaning	concave up	minimum turning point	concave up

As shown in the table, as there is no change in concavity at $x=0$x=0 there isn't a point of inflection at this location.

This example highlights the importance of checking concavity either side of points where $f''(x)=0$f′′(x)=0 . The first derivative of this function told us that there was a stationary point at $x=0$x=0. However, in this case, we still needed to look at the sign of the second derivative left and right of the point to determine if there was a change in concavity.

Summary

The following table summarises our first derivative and second derivative curve analysis.

	$f'\left(x\right)<0$f′(x)<0	$f'\left(x\right)=0$f′(x)=0	$f'\left(x\right)>0$f′(x)>0
$f''\left(x\right)<0$f′′(x)<0
$f''\left(x\right)=0$f′′(x)=0
$f''\left(x\right)>0$f′′(x)>0

Practice questions

Question 2

Question 3

Question 4