Second Order Derivatives (Cambridge O Level Additional Maths)

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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 fraction numerator straight d squared y over denominator straight d x squared end fraction
  • Note the position of the powers of 2
    • differentiating twice (sobold italic space bold d to the power of bold 2) with respect to x twice (sobold space bold italic x to the power of bold 2)
  • A first derivative is the rate of change of a function (the gradient)
    • 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 equals 5 x cubed plus 10 x squared
      • fraction numerator d y over denominator d x end fraction equals 15 x squared plus 20 x
      • fraction numerator straight d squared y over denominator straight d x squared end fraction equals 30 x plus 20

Examiner Tip

  • 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 fraction numerator straight d squared y over denominator straight d x squared end fraction when

(a)
y equals x to the power of 5 minus 2 x cubed plus 7 x squared plus 9 x minus 18
 

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

table row cell fraction numerator d y over denominator d x end fraction end cell equals cell space 5 x to the power of 4 space minus space 6 x squared space plus space 14 x space plus space 9 end cell end table 

Find the second derivative, fraction numerator d squared y over denominator d x squared end fraction by differentiating each term in the first derivative.

table row cell fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction end cell bold equals cell bold space bold 20 bold italic x to the power of bold 3 bold space bold minus bold space bold 12 bold italic x bold space bold plus bold space bold 14 end cell end table

 

(b)table row blank row blank row blank end table
y equals fraction numerator 3 x plus 7 over denominator x to the power of 4 end fraction
 

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

table row cell y space end cell equals cell space fraction numerator 3 x over denominator x to the power of 4 end fraction space plus space 7 over x to the power of 4 end cell end table  

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

table row cell y space end cell equals cell space 3 x open parentheses x to the power of negative 4 end exponent close parentheses space plus space 7 x to the power of negative 4 end exponent space equals space 3 x to the power of negative 3 end exponent space plus space 7 x to the power of negative 4 end exponent end cell end table 

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

table row cell fraction numerator d y over denominator d x end fraction end cell equals cell space minus 9 x to the power of negative 4 end exponent space minus space 28 x to the power of negative 5 end exponent end cell end table 

Find the second derivative, fraction numerator d squared y over denominator d x squared end fraction by differentiating each term in the first derivative.

table row cell fraction numerator d squared y over denominator d x squared end fraction end cell equals cell space minus 9 open parentheses negative 4 close parentheses x to the power of negative 5 end exponent space minus space 28 open parentheses negative 5 close parentheses x to the power of negative 6 end exponent space equals space 36 x to the power of negative 5 end exponent space plus space 140 x to the power of negative 6 end exponent end cell end table

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

table row cell fraction numerator d squared y over denominator d x squared end fraction end cell equals cell space 36 over x to the power of 5 plus space 140 over x to the power of 6 space equals space fraction numerator 36 x space plus space 140 over denominator x to the power of 6 end fraction end cell end table

table row cell fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction end cell bold equals cell bold space fraction numerator bold 36 bold italic x bold space bold plus bold space bold 140 over denominator bold italic x to the power of bold 6 end fraction end cell end table

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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 fraction numerator d y over denominator d x end fraction

Step 2
Solve fraction numerator d y over denominator d x end fraction equals 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 equals x squared space minus space 2 x

and explain why it will be a minimum point.

Find the derivative.

fraction numerator straight d y over denominator straight d x end fraction equals 2 x space minus space 2

Solve fraction numerator straight d y over denominator straight d x end fraction equals 0

table row cell 2 x space minus space 2 space end cell equals cell space 0 end cell row cell 2 x space end cell equals cell space 2 end cell row cell x space end cell equals cell space 1 end cell end table


Find the corresponding y-coordinate.

table row x equals cell 1 comma space space y equals open parentheses 1 close parentheses squared space minus space 2 open parentheses 1 close parentheses space equals space minus 1 end cell end table

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, fraction numerator straight d y over denominator straight d x end fraction, (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 fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction
    • For each stationary point find the value of fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction at the stationary point
      • i.e. substitute the x-coordinate of the stationary point into fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction and evaluate
    • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction is positive then the point is a local minimum
    • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction is negative then the point is a local maximum
    • If fraction numerator bold d to the power of bold 2 bold italic y over denominator bold d bold italic x to the power of bold 2 end fraction 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 Tip

  • 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 equals x squared open parentheses 2 x squared minus 4 close parentheses

and determine the nature of each.

Start by expanding the brackets in the expression for y

y equals 2 x to the power of 4 minus 4 x squared

Step 1
Find the first derivative

fraction numerator straight d y over denominator straight d x end fraction equals 8 x cubed minus 8 x

Step 2
Solve fraction numerator straight d y over denominator straight d x end fraction equals 0

table row cell 8 x cubed minus 8 x end cell equals 0 end table

Divide through by 8

table attributes columnalign right center left columnspacing 0px end attributes row cell x cubed minus x end cell equals 0 end table

Fully factorise, spotting the difference of two squares

table row cell x open parentheses x squared minus 1 close parentheses end cell equals 0 row cell x open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end cell equals 0 end table

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

x subscript 1 space equals space 0 comma space x subscript 2 space equals space 1 comma space x subscript 3 space equals space minus 1

Step 3
Find the corresponding y-coordinates

table row cell x subscript 1 end cell equals cell 0 comma space space space space space space space space y subscript 1 equals open parentheses 0 close parentheses squared open parentheses 2 open parentheses 0 close parentheses squared minus 4 close parentheses equals 0 end cell row cell x subscript 2 end cell equals cell 1 comma space space space space space space space space y subscript 2 equals open parentheses 1 close parentheses squared open parentheses 2 open parentheses 1 close parentheses squared minus 4 close parentheses equals negative 2 end cell row cell x subscript 3 end cell equals cell negative 1 comma space space space space y subscript 3 equals open parentheses negative 1 close parentheses squared open parentheses 2 open parentheses negative 1 close parentheses squared minus 4 close parentheses equals negative 2 end cell end table

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

fraction numerator straight d squared y over denominator straight d x squared end fraction equals 24 x squared minus 8

Step 5
Evaluate the second derivative at each stationary point

table row x equals cell 0 comma fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 24 open parentheses 0 close parentheses squared minus 8 equals negative 8 space space open parentheses less than 0 close parentheses end cell row x equals cell 1 comma space fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 24 open parentheses 1 close parentheses squared minus 8 equals 16 space space space open parentheses greater than 0 close parentheses end cell row x equals cell negative 1 comma space fraction numerator space straight d squared y over denominator straight d x squared end fraction equals 24 open parentheses negative 1 close parentheses squared minus 8 equals 16 space open parentheses greater than 0 close parentheses end cell end table

The nature of each stationary point is now determined

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

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Amber

Author: Amber

Expertise: Maths

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.