Differentiating Basic Functions (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Differentiating Powers of x

What is differentiation?

  • Differentiation is the process of finding the derivative (gradient function) of a function

How do I differentiate powers of x?

  • Powers of bold italic x are differentiated according to the following formula:

    • Ifspace straight f left parenthesis x right parenthesis equals x to the power of n thenspace straight f apostrophe left parenthesis x right parenthesis equals n x to the power of n minus 1 end exponent

      • Bring the power down in front as a multiplier

      • Then subtract 1 from the power

    • This formula is not on the exam formula sheet, so you need to remember it

  • If the power of x term is multiplied by a constant a

    • then the derivative is also multiplied by that constant

      • If space straight f left parenthesis x right parenthesis equals a x to the power of n  then straight f to the power of apostrophe left parenthesis x right parenthesis equals a n x to the power of n minus 1 end exponent

  • The alternative notation (tospace straight f to the power of apostrophe left parenthesis x right parenthesis) is to use fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction

    • 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

  • Don't forget these two special cases:

    • Ifspace straight f left parenthesis x right parenthesis equals a x thenspace straight f to the power of apostrophe left parenthesis x right parenthesis equals a

      • e.g.  If y equals 6 x then fraction numerator straight d y over denominator straight d x end fraction equals 6

    • Ifspace straight f left parenthesis x right parenthesis equals a thenspace straight f to the power of apostrophe left parenthesis x right parenthesis equals 0

      • e.g.  If y equals 5  (or if y equals any constant) then fraction numerator straight d y over denominator straight d x end fraction equals 0

  • Functions involving roots will need to be rewritten as fractional powers

    • e.g.  straight f left parenthesis x right parenthesis equals 2 square root of x

      • rewrite asspace straight f left parenthesis x right parenthesis equals 2 x to the power of 1 half end exponent

      • then differentiate

  • Functions involving fractions with x in the denominator will need to be rewritten as negative powers

    • e.g.  straight f left parenthesis x right parenthesis equals 4 over x

      • rewrite asspace straight f left parenthesis x right parenthesis equals 4 x to the power of negative 1 end exponent

      • then differentiate

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

  • The formulae can be used to differentiate sums or differences of powers of x

    • Just differentiate term by term 

      • e.g. space straight f left parenthesis x right parenthesis equals 5 x to the power of 4 minus 3 x to the power of 2 over 3 end exponent plus 4

      • table row cell straight f apostrophe left parenthesis x right parenthesis end cell equals cell 5 cross times 4 x to the power of 4 minus 1 end exponent minus 3 cross times 2 over 3 x to the power of 2 over 3 minus 1 end exponent plus 0 end cell end table

      • table row cell straight f apostrophe left parenthesis x right parenthesis end cell equals cell 20 x cubed minus 2 x to the power of negative 1 third end exponent end cell end table

  • Products and quotients cannot be differentiated in this way

    • These need to be expanded/simplifying first

      • e.g.  straight f left parenthesis x right parenthesis equals left parenthesis 2 x minus 3 right parenthesis left parenthesis x squared minus 4 right parenthesis

      • Expand tospace straight f left parenthesis x right parenthesis equals 2 x cubed minus 3 x squared minus 8 x plus 12

      • Then differentiate term by term

      • You can't just multiply the derivatives of 2 x minus 3 and x squared minus 4 together!

    • These can also be differentiated using the product rule or quotient rule

Examiner Tips and Tricks

  • Be careful with negative and fractional powers

  • It's easy to make a mistake when subtracting 1 from these

Worked Example

The functionspace straight f left parenthesis x right parenthesis is given by

space straight f left parenthesis x right parenthesis equals 2 x cubed plus fraction numerator 4 over denominator square root of x end fraction,  where x greater than 0

Find the derivative of straight f open parentheses x close parentheses.


Start by rewriting the fraction numerator 4 over denominator square root of x end fraction term as a power of x

By laws of indices, fraction numerator 1 over denominator square root of x end fraction equals x to the power of negative 1 half end exponent

straight f open parentheses x close parentheses equals 2 x cubed plus 4 x to the power of negative 1 half end exponent

Now differentiate as powers of x

table row cell straight f to the power of apostrophe open parentheses x close parentheses end cell equals cell 2 open parentheses 3 x to the power of 3 minus 1 end exponent close parentheses plus 4 open parentheses negative 1 half x to the power of negative 1 half minus 1 end exponent close parentheses end cell row blank equals cell 2 open parentheses 3 x squared close parentheses plus 4 open parentheses negative 1 half x to the power of negative 3 over 2 end exponent close parentheses end cell row blank equals cell 6 x squared minus 2 x to the power of negative 3 over 2 end exponent end cell end table

bold f to the power of bold apostrophe stretchy left parenthesis x stretchy right parenthesis bold equals bold 6 bold italic x to the power of bold 2 bold minus bold 2 bold italic x to the power of bold minus bold 3 over bold 2 end exponent

Differentiating Trig Functions

How do I differentiate sin and cos?

  • The derivative of  bold italic y bold equals bold sin bold italic x  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold cos bold italic x  

  • The derivative of is  bold italic y bold equals bold cos bold italic x  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold minus bold sin bold italic x

  • If x is multiplied by a constant a then

    • the derivative of  bold italic y bold equals bold sin bold italic a bold italic x  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold italic a bold cos bold italic a bold italic x

    • the derivative of  bold italic y bold equals bold cos bold italic a bold italic x  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold minus bold italic a bold sin bold italic a bold italic x

      • These can be derived by using the chain rule

      • but it's easier (and quicker) just to remember them

  • None of these formulae are on the exam formula sheet, so you need to remember them

  • For calculus with trigonometric functions angles must be measured in radians

    • Make sure you know how to change the angle mode on your calculator

Examiner Tips and Tricks

  • As soon as you see a question involving differentiation and trigonometry

    • put your calculator into radians mode

Worked Example

(a) Given the function straight f open parentheses x close parentheses equals cos 5 x, find space straight f to the power of apostrophe left parenthesis x right parenthesis.

Use  y equals cos a x space rightwards double arrow space fraction numerator straight d y over denominator straight d x end fraction equals negative a sin a x  with  a equals 5

bold f to the power of bold apostrophe begin bold style stretchy left parenthesis x stretchy right parenthesis end style bold equals bold minus bold 5 bold sin bold 5 bold italic x

 

(b) A curve has the equation y equals 3 sin x over 2.
Find the gradient of the curve at the point where  x equals pi over 3, giving your answer as an exact value.


Start by finding fraction numerator straight d y over denominator straight d x end fraction

Use  y equals sin a x space rightwards double arrow space fraction numerator straight d y over denominator straight d x end fraction equals a cos a x  with  a equals 1 half

fraction numerator straight d y over denominator straight d x end fraction equals 3 open parentheses 1 half cos x over 2 close parentheses equals 3 over 2 cos x over 2


Substitute x equals pi over 3 into fraction numerator straight d y over denominator straight d x end fraction to find the gradient

gradient equals 3 over 2 cos open parentheses fraction numerator open parentheses pi over 3 close parentheses over denominator 2 end fraction close parentheses equals 3 over 2 cos pi over 6 equals 3 over 2 open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses equals fraction numerator 3 square root of 3 over denominator 4 end fraction


fraction numerator bold 3 square root of bold 3 over denominator bold 4 end fraction

Differentiating e^x

How do I differentiate exponentials?

  • The derivative of  bold italic y bold equals bold e to the power of bold italic x  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold e to the power of bold italic x

    • Note that straight e to the power of x is its own derivative!

  • If x is multiplied by a constant a then

    • the derivative of  bold italic y bold equals bold e to the power of bold italic a bold italic x end exponent  is  fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold italic a bold e to the power of bold italic a bold italic x end exponent

      • This can be derived by using the chain rule

      • but it's easier (and quicker) just to remember it

Examiner Tips and Tricks

  • Remember this is not a 'powers of x' differentiation

    • the derivative of straight e to the power of k x end exponent is k straight e to the power of k x end exponent, NOT k x straight e to the power of k x minus 1 end exponent

Worked Example

A curve has the equationspace y equals 2 straight e to the power of negative 3 x end exponent.

Find the gradient of the curve at the point wherespace x equals 1 half, giving your answer correct to 3 significant figures.

Differentiate using y equals straight e to the power of a x end exponent space rightwards double arrow space fraction numerator straight d y over denominator straight d x end fraction equals a straight e to the power of a x end exponent

fraction numerator straight d y over denominator straight d x end fraction equals 2 open parentheses negative 3 straight e to the power of negative 3 x end exponent close parentheses equals negative 6 straight e to the power of negative 3 x end exponent


Substitute x equals 1 half into fraction numerator straight d y over denominator straight d x end fraction to find the gradient

gradient equals negative 6 straight e to the power of negative 3 open parentheses 1 half close parentheses end exponent equals negative 6 straight e to the power of negative 3 over 2 end exponent


Note that  negative 6 straight e to the power of negative 3 over 2 end exponent equals negative fraction numerator 6 over denominator straight e square root of straight e end fraction  is the exact value answer

Use a calculator to find the decimal version

negative 6 straight e to the power of negative 3 over 2 end exponent equals negative 1.338780...


bold minus bold 1 bold. bold 34 bold space bold space begin bold style stretchy left parenthesis 3 space s. f. stretchy right parenthesis end style

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

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.