Introduction to Differentiation (Cambridge O Level Additional Maths)

Revision Note

Test yourself
Amber

Author

Amber

Last updated

Did this video help you?

Definition of Gradient

What is the gradient of a curve?

  • At a given point the gradient of a curve is defined as the gradient of the tangent to the curve at that point
  • A tangent to a curve is a line that just touches the curve at one point but doesn't cut the curve at that point

Def Grad Illustr 1, A Level & AS Maths: Pure revision notes

  • A tangent may cut the curve somewhere else on the curve

Def Grad Illustr 2, A Level & AS Maths: Pure revision notes

  • It is only possible to draw one tangent to a curve at any given point
  • Note that unlike the gradient of a straight line, the gradient of a curve is constantly changing

Examiner Tip

  • If a question asks for the "rate of change of ..." then it is asking for the "gradient"

Worked example

The diagram shows the curve with equation y equals x cubed minus 2 x squared minus x plus 3. The tangent, T, to the curve at the point A left parenthesis 2 comma space 1 right parenthesis is also shown.

definition-of-gradient-we

Using the diagram, calculate the gradient of the curve at A.

The gradient of the curve at the point A is the same as the gradient of the tangent T.
Calculate the gradient of the line.

definition-of-gradient-ma

fraction numerator 4 minus open parentheses negative 2 close parentheses over denominator 3 minus 1 end fraction

The gradient is 3

Definition of Derivatives

What is a derivative?

  • Calculus is about rates of change
    • the way a car’s position on a road changes is its speed (velocity)
    • the way the car’s speed changes is its acceleration
  • The gradient (rate of change) of a (non-linear) function varies with x
  • The derivative of a function is a function that relates the gradient to the value of x
    • For example, the derivative of  y equals x squared  is  2 x
      • This means that when x equals 1, the gradient of y equals x squared  is  2 open parentheses 1 close parentheses equals 2
      • And when x equals 5, the gradient of y equals x squared  is  2 open parentheses 5 close parentheses equals 10
  • The derivative is also called the gradient function

Worked example

The derivative of y equals x cubed minus 2 x squared minus x plus 3 is 3 x squared minus 4 x minus 1.

Use the derivative to find the gradient of y equals x cubed minus 2 x squared minus x plus 3 at the point A open parentheses 2 comma space 1 close parentheses.

Substitute x equals 2 into the derivative, 3 x squared minus 4 x minus 1

3 open parentheses 2 close parentheses squared minus 4 open parentheses 2 close parentheses minus 1 equals 12 minus 8 minus 1

gradient = 3

Note that the answer is the same as in the method above

Did this video help you?

Differentiating Powers of x

What is differentiation?

  • Differentiation is the process of finding an expression for the derivative (gradient function) from the equation of a curve
    • The equation of the curve is written y equals... and the gradient function is written fraction numerator straight d y over denominator straight d x end fraction equals...

How do I differentiate powers of x?

  • Powers of x are differentiated according to the following formula:
    • If y equals a x to the power of n then fraction numerator straight d y over denominator straight d x end fraction equals a n x to the power of n minus 1 end exponent
      • e.g.  If y equals 4 x cubed then fraction numerator straight d y over denominator straight d x end fraction equals 4 cross times 3 cross times x to the power of 3 minus 1 end exponent equals 12 x squared
      • you "bring down the power" then "subtract one from the power"
  • Don't forget these two special cases:
    • If y equals a x thenfraction numerator straight d y over denominator straight d x end fraction equals a
      • e.g.  If y equals 6 x then fraction numerator straight d y over denominator straight d x end fraction equals 6
    • If y equals a thenfraction numerator straight d y over denominator straight d x end fraction equals 0
      • e.g.  If y equals 5 then fraction numerator straight d y over denominator straight d x end fraction equals 0
    • These allow you to differentiate linear terms in x and constants
  • Functions involving fractions with denominators in terms of x will need to be rewritten as negative powers of x first
    • e.g.  If y equals 4 over x then rewrite as y equals 4 x to the power of negative 1 end exponent and differentiate

How do I differentiate sums and differences of powers of x?

  •  The formulae for differentiating powers of x work for a sum or difference of powers of x
    • e.g.  Ifspace y equals 5 x to the power of 4 plus 3 x to the power of negative 2 end exponent plus 4 then
      fraction numerator d y over denominator d x end fraction equals 5 cross times 4 x to the power of 4 minus 1 end exponent plus 3 cross times open parentheses negative 2 close parentheses x to the power of negative 2 minus 1 end exponent plus 0
      fraction numerator d y over denominator d x end fraction equals 20 x cubed minus 6 x to the power of negative 3 end exponent
    • This is sometimes referred to differentiating 'term-by-term'
  • Products and quotients (divisions) cannot be differentiated in this way so they need expanding/simplifying first
    • e.g.  If y equals left parenthesis 2 x minus 3 right parenthesis left parenthesis x squared minus 4 right parenthesis then expand to y equals 2 x cubed minus 3 x squared minus 8 x plus 12 which is a sum/difference of powers of x and can then be differentiated

What can I do with derivatives (gradient functions)?

  • The derivative can be used to find the gradient of a function at any point
    • The gradient of a function at a point is equal to the gradient of the tangent to the curve at that point
    • A question may refer to the gradient of the tangent

Examiner Tip

  • Don't try to do too many steps in your head; write the expression in a format that you can differentiate before you actually differentiate it
    • e.g. y equals 1 over x to the power of 4 plus 2 over x cubed can be rewritten as y equals x to the power of negative 4 end exponent plus 2 x to the power of negative 3 end exponent which is then far easier to differentiate

Worked example

Find the derivative of 

(a)table row blank row blank row blank end table

y equals 5 x cubed plus 2 x plus 3 over x squared plus 8

Rewrite the 3 over x squared term

y equals 5 x cubed plus 2 x plus 3 x to the power of negative 2 end exponent plus 8

Apply the rule for differentiating powers (y equals a x to the power of n comma space fraction numerator straight d y over denominator straight d x end fraction equals a n x to the power of n minus 1 end exponent) and apply the special cases for the terms 2 x and 8 (y equals a x comma fraction numerator space straight d y over denominator straight d x end fraction equals a and y equals a comma space fraction numerator straight d y over denominator straight d x end fraction equals 0)

fraction numerator straight d y over denominator straight d x end fraction equals 15 x squared plus 2 minus 6 x to the power of negative 3 end exponent

Unless a question specifies there is not usually a need to rewrite/simplify the answer

fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold 15 bold italic x to the power of bold 2 bold plus bold 2 bold minus bold 6 over bold italic x to the power of bold 3

 
(b)

y equals open parentheses 2 x plus 3 close parentheses squared

This is a product of two (equal) brackets so cannot be differentiated directly
Expand the brackets to get an expression in powers of x
Take time to get the expansion correct, writing stages out in full if necessary

table row y equals cell open parentheses 2 x plus 3 close parentheses open parentheses 2 x plus 3 close parentheses end cell row y equals cell 4 x squared plus 6 x plus 6 x plus 9 end cell row y equals cell 4 x squared plus 12 x plus 9 end cell end table

Differentiate 'term-by-term', looking out for those special cases

fraction numerator straight d y over denominator straight d x end fraction equals 8 x plus 12

There is a factor of 4 but there is no demand to factorise the final answer in the question

fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold 8 bold italic x bold plus bold 12

 
(c)begin mathsize 12px style table row blank row blank row blank end table end style

y equals fraction numerator 8 x to the power of 6 minus x cubed over denominator 2 x to the power of 4 end fraction

This is a quotient so cannot be differentiated directly
Spot the single denominator which means we can split the fraction by the two terms on the numerator

y equals fraction numerator 8 x to the power of 6 over denominator 2 x to the power of 4 end fraction minus fraction numerator x cubed over denominator 2 x to the power of 4 end fraction

Simplify using the laws of indices

y equals 4 x to the power of 6 minus 4 end exponent minus 1 half x to the power of 3 minus 4 end exponent
y equals 4 x squared minus 1 half x to the power of negative 1 end exponent

Each term is now a power of x, so differentiate 'term-by-term'

fraction numerator straight d y over denominator straight d x end fraction equals 8 x plus 1 half x to the power of negative 2 end exponent

There is demand to simplify or write the answer in a particular form

fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold 8 bold italic x bold plus bold 1 over bold 2 bold italic x to the power of bold minus bold 2 end exponent

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

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.