Second Order Derivatives (Cambridge (CIE) O Level Additional Maths): Revision Note

Exam code: 4037

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Second order derivatives

What is the second order derivative of a function?

  • If you differentiate the derivative of a function (i.e. differentiate the function a second time) you get the second order derivative of the function

    • The second order derivative can be referred to simply as the second derivative

  • We can write the second derivative as d2ydx2

  • Note the position of the powers of 2

    • differentiating twice (so d2) with respect to x twice (so x2)

  • A first derivative is the rate of change of a function (the gradient)

    • a second order derivative is the rate of change of the rate of change of a function

      • i.e. the rate of change of the function’s gradient

    • A positive second derivative means the gradient is increasing

      • For instance in a u-shape, the gradient is changing from negative to positive

    • A negative second derivative means the gradient is decreasing

      • For instance in an n-shape, the gradient is changing from positive to negative

  • Second order derivatives can be used to test whether a point is a minimum or maximum

  • To find a second derivative, you simply differentiate twice!

    • It is important to write down your working with the correct notation, so you know what each expression means

    • For example

      • y=5x3+10x2

      • dydx=15x2+20x

      • d2ydx2=30x+20

Examiner Tips and Tricks

  • Even if you think you can find the second derivative in your head and write it down, make sure you write down the first derivative as well

    • If you make a mistake, you will most likely get marks for finding the first derivative

Worked Example

Work out d2ydx2 when

(a) y=x5−2x3+7x2+9x−18

 

Find the first derivative of the function first by considering each term in turn.

dydx= 5x4 − 6x2 + 14x + 9 

Find the second derivative, d2ydx2 by differentiating each term in the first derivative.

d2ydx2= 20x3 − 12x + 14  

(b) y=3x+7x4

 

To find the first derivative of the function, begin by separating the terms in the fraction.

y = 3xx4 + 7x4  

Rewrite each term using index notation so that they are in a form that can be differentiated. 

y = 3x(x−4) + 7x−4 = 3x−3 + 7x−4 

Find the first derivative of the function first by differentiating each term in turn.

dydx= −9x−4 − 28x−5 

Find the second derivative, d2ydx2 by differentiating each term in the first derivative.

d2ydx2= −9(−4)x−5 − 28(−5)x−6 = 36x−5 + 140x−6

You can turn the second derivative back into the same format as the original function by rewriting as a fraction.

d2ydx2= 36x5+ 140x6 = 36x + 140x6

d2ydx2= 36x + 140x6

Stationary points & turning points

What are stationary points?

  • A stationary point is any point on a curve where the gradient is zero

  • To find stationary points of a curve

Step 1 Find the first derivative dydx

Step 2 Solve dydx=0 to find the x-coordinates of any stationary points

Step 3 Substitute those x-coordinates into the equation of the curve to find the corresponding y-coordinates

  • A stationary point may be either a local minimum, a local maximum, or a point of inflection

 

A stationary point is a local minimum, local maximum or a point of inflection

Stationary points on quadratics

  • The graph of a quadratic function only has a single stationary point

  • For a positive quadratic this is the minimum; for a negative quadratic it is the maximum

    • No need to talk about 'local' here, as it is the overall minimum/maximum for the whole curve

Stationary points on quadratic graphs
  • The y value/coordinate of the stationary point is therefore the minimum or maximum value of the quadratic function

  • For quadratics especially, minimum and maximum points are often referred to as turning points

Worked Example

Find the stationary point of

y=x2 − 2x

and explain why it will be a minimum point.

Find the derivative.

dydx=2x − 2

Solve dydx=0

2x − 2 = 02x = 2x = 1

Find the corresponding y-coordinate.

x=1,  y=(1)2 − 2(1) = −1

Write the answer as a coordinate.

The stationary point is (1, -1)

To explain why it is a minimum point, consider the shape of the quadratic. 

(1, -1) is a minimum point because it is a positive quadratic.

Testing for local minimum & maximum points

How do I determine the nature of stationary points on a curve?

  • For a graph there are two ways to determine the nature of its stationary points 

  • Method A

    • Compare the signs of the first derivative, dydx, (positive or negative) a little bit to either side of the stationary point

      • e.g. if the stationary point is at x=2 then you could find the gradient at x=1.9 and x=2.1

  • Compare the signs (positive or negative) of the derivatives on the left and right of the stationary point

    • If the derivatives are negative on the left and positive on the right, the point is a local minimum (a u-shape)

    • If the derivatives are positive on the left and negative on the right, the point is a local maximum (an n-shape)

The gradient of a curve either side of a stationary point

 

  • Method B

    • Look at the sign of the second derivative (positive or negative) at the stationary point

    • Find the second derivative d2ydx2

    • For each stationary point find the value of d2ydx2 at the stationary point

      • i.e. substitute the x-coordinate of the stationary point into d2ydx2 and evaluate

  • If d2ydx2 is positive then the point is a local minimum

    • If d2ydx2 is negative then the point is a local maximum

    • If d2ydx2 is zero then the point could be a local minimum, a local maximum OR a point of inflection

      • In this case you will need to use method A instead

Examiner Tips and Tricks

  • Usually, using the second derivative (Method B above) is a much quicker way of determining the nature of a stationary point

  • Sketching the curve can also help

Worked Example

Find the stationary points of

y=x2(2x2−4)

and determine the nature of each.

Start by expanding the brackets in the expression for y

y=2x4−4x2

Step 1 Find the first derivative

dydx=8x3−8x

Step 2 Solve dydx=0

8x3−8x=0

Divide through by 8

x3−x=0

Fully factorise, spotting the difference of two squares

x(x2−1)=0x(x−1)(x+1)=0

Solve to find the x-coordinates of the stationary (turning) points

x1 = 0, x2 = 1, x3 = −1

Step 3 Find the corresponding y-coordinates

x1=0,        y1=(0)2(2(0)2−4)=0x2=1,        y2=(1)2(2(1)2−4)=−2x3=−1,    y3=(−1)2(2(−1)2−4)=−2

Write the answers as coordinates, being careful to correctly match x's and y's

The stationary points are (0, 0), (1, -2) and (-1, -2)

To find the nature of each stationary point, use Method B

Step 4 Find the second derivative

d2ydx2=24x2−8

Step 5 Evaluate the second derivative at each stationary point

x=0, d2ydx2=24(0)2−8=−8  (<0)x=1,  d2ydx2=24(1)2−8=16   (>0)x=−1,  d2ydx2=24(−1)2−8=16 (>0)

The nature of each stationary point is now determined

(0, 0) is a local maximum point
(1, -2) is a local minimum point
(-1, -2) is a local minimum point

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Amber

Author: Amber

Expertise: Maths Content Creator

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Dan Finlay

Reviewer: Dan Finlay

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Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.