Techniques of Differentiation (DP IB Maths: AI HL)

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Lucy

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Lucy

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Chain Rule

What is the chain rule?

  •  The chain rule states ifbold space bold italic y is a function ofbold space bold italic u andbold space bold italic u is a function ofbold space bold italic x then

space y equals f left parenthesis u left parenthesis x right parenthesis right parenthesis

space fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals fraction numerator bold d bold italic y over denominator bold d bold italic u end fraction bold cross times fraction numerator bold d bold italic u over denominator bold d bold italic x end fraction

    • This is given in the formula booklet

  • In function notation this could be written

space y equals f left parenthesis g left parenthesis x right parenthesis right parenthesis

space fraction numerator straight d y over denominator straight d x end fraction equals f apostrophe left parenthesis g left parenthesis x right parenthesis right parenthesis g apostrophe left parenthesis x right parenthesis

How do I know when to use the chain rule?

  •  The chain rule is used when we are trying to differentiate composite functions
    • “function of a function”
    • these can be identified as the variable (usuallyspace x) does not ‘appear alone’
      • space sin space x – not a composite function, x ‘appears alone’
      • sin left parenthesis 3 x plus 2 right parenthesis is a composite function; x is tripled and has 2 added to it before the sine function is applied

How do I use the chain rule?

 STEP 1
 Identify the two functions
 Rewrite y as a function ofspace u; space y equals f left parenthesis u right parenthesis
 Write u as a function ofspace xspace u equals g left parenthesis x right parenthesis

 STEP 2
Differentiate y with respect to u to getspace fraction numerator straight d y over denominator straight d u end fraction
Differentiate u with respect to x to getspace fraction numerator straight d u over denominator straight d x end fraction

 STEP 3
Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the formulaspace fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d y over denominator straight d u end fraction cross times fraction numerator straight d u over denominator straight d x end fraction and substitutespace u back in forspace g left parenthesis x right parenthesis
 
  • In trickier problems chain rule may have to be applied more than once

Are there any standard results for using chain rule?

  • There are five general results that can be useful
    • Ifsize 16px space size 16px y size 16px equals begin mathsize 16px style stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis end style to the power of size 16px n thenspace fraction numerator straight d y over denominator straight d x end fraction equals n straight f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis straight f stretchy left parenthesis x stretchy right parenthesis to the power of n minus 1 end exponent

    • Ifspace y equals straight e to the power of space f stretchy left parenthesis x stretchy right parenthesis end exponent thenspace fraction numerator straight d y over denominator straight d x end fraction equals f to the power of apostrophe left parenthesis x right parenthesis e to the power of space f stretchy left parenthesis x stretchy right parenthesis end exponent
    • Ifspace y equals ln stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis thenspace fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis over denominator f stretchy left parenthesis x stretchy right parenthesis end fraction
    • Ifsize 16px space size 16px y size 16px equals size 16px sin begin mathsize 16px style stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis end style thenspace fraction numerator straight d y over denominator straight d x end fraction equals f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis cos stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis
    • Ifspace y equals cos stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis thenspace fraction numerator straight d y over denominator straight d x end fraction equals negative f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis sin stretchy left parenthesis f open parentheses x close parentheses stretchy right parenthesis

Examiner Tip

  • You should aim to be able to spot and carry out the chain rule mentally (rather than use substitution)
    • every time you use it, say it to yourself in your head
      “differentiate the first function ignoring the second, then multiply by the derivative of the second function"

Worked example

a)
Find the derivative ofspace y equals left parenthesis x squared minus 5 x plus 7 right parenthesis to the power of 7.

5-2-2-ib-sl-aa-only-chain-we-soltn-a

b)
Find the derivative ofspace y equals sin left parenthesis straight e to the power of 2 x end exponent right parenthesis.

5-2-2-ib-sl-aa-only-chain-we-soltn-b

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Product Rule

What is the product rule?

  • The product rule states ifspace y is the product of two functionsspace u left parenthesis x right parenthesis andspace v left parenthesis x right parenthesis then

space y equals u v

bold space fraction numerator bold d bold italic y over denominator bold d bold italic x end fraction bold equals bold italic u fraction numerator bold d bold italic v over denominator bold d bold italic x end fraction bold plus bold italic v fraction numerator bold d bold italic u over denominator bold d bold italic x end fraction 

    • This is given in the formula booklet
  • In function notation this could be written as
y equals f left parenthesis x right parenthesis g left parenthesis x right parenthesis
space fraction numerator straight d y over denominator straight d x end fraction equals f left parenthesis x right parenthesis g to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis plus g left parenthesis x right parenthesis f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis

  • ‘Dash notation’ may be used as a shorter way of writing the rule
space y equals u v
size 16px space size 16px y to the power of size 16px apostrophe size 16px equals size 16px u size 16px v to the power of size 16px apostrophe size 16px plus size 16px v size 16px u size 16px apostrophe
  • Final answers should match the notation used throughout the question

How do I know when to use the product rule?

  • The product rule is used when we are trying to differentiate the product of two functions
    • these can easily be confused with composite functions (see chain rule)
      • space sin left parenthesis cos space x right parenthesis is a composite function, “sin of cos of x
      •  space sin space x cos space x is a product, “sin x times cos x

How do I use the product rule?

  • Make it clear whatspace u comma space v comma space u apostrophe andspace v apostrophe are
    • arranging them in a square can help
      • opposite diagonals match up

 STEP 1
 Identify the two functions,space u andspace v
 Differentiate bothspace u andspace v with respect tospace x to findspace u apostrophe and v apostrophe

 STEP 2
Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the product rule formula
Simplify the answer if straightforward to do so or if the question requires a particular form

  • In trickier problems chain rule may have to be used when finding u apostrophe and v apostrophe

Examiner Tip

  • Use u comma space v comma space u apostrophe and v apostrophe for the elements of product rule
    • lay them out in a 'square' (imagine a 2x2 grid)
    • those that are paired together are then on opposite diagonals (u and v apostrophev and u apostrophe)
  • For trickier functions chain rule may be required inside product rule
    • i.e.  chain rule may be needed to differentiate u and v

Worked example

a)       Find the derivative ofspace y equals straight e to the power of x sin space x.

5-2-2-ib-sl-aa-only-product-we-soltn-a

b)       Find the derivative ofsize 16px space size 16px y size 16px equals size 16px 5 size 16px x to the power of size 16px 2 size 16px space size 16px cos size 16px space size 16px 3 size 16px x to the power of size 16px 2.

5-2-2-ib-sl-aa-only-product-we-soltn-a-b

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Quotient Rule

What is the quotient rule?

  • The quotient rule states if space y is the quotient fraction numerator u left parenthesis x right parenthesis over denominator v left parenthesis x right parenthesis end fraction then

space y equals u over v

 

    • This is given in the formula booklet
  • In function notation this could be written
space y equals fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction

  • As with product rule, ‘dash notation’ may be used
y equals u over v
y apostrophe equals fraction numerator v u apostrophe minus u v apostrophe over denominator v squared end fraction
  • Final answers should match the notation used throughout the question

How do I know when to use the quotient rule?

  • The quotient rule is used when trying to differentiate a fraction where both the numerator and denominator are functions ofspace x
    • if the numerator is a constant, negative powers can be used
    • if the denominator is a constant, treat it as a factor of the expression

How do I use the quotient rule?

  • Make it clear whatspace u comma space v comma space u apostrophe andspace v apostrophe are
    • arranging them in a square can help
      • opposite diagonals match up (like they do for product rule)

 STEP 1
 Identify the two functions,space u andspace v
 Differentiate bothspace u andspace v with respect tospace x to findspace u apostrophe andspace v apostrophe

 STEP 2
Obtain fraction numerator straight d y over denominator straight d x end fraction by applying the quotient rule formula
Be careful using the formula – because of the minus sign in the numerator, the order of the functions is important
Simplify the answer if straightforward or if the question requires a particular form

  • In trickier problems chain rule may have to be used when findingspace u apostrophe andspace v apostrophe,

Examiner Tip

  • Use u comma space v comma space u apostrophe and v apostrophe for the elements of quotient rule
    • lay them out in a 'square' (imagine a 2x2 grid)
    • those that are paired together are then on opposite diagonals (v and u apostrophe,  u and v apostrophe)
  • Look out for functions of the form space y equals f left parenthesis x right parenthesis left parenthesis g left parenthesis x right parenthesis right parenthesis to the power of negative 1 end exponent
    • These can be differentiated using a combination of chain rule and product rule
      (it would be good practice to try!)
    • ... but it can also be seen as a quotient rule question in disguise 
    • ... and vice versa!
      • A quotient could be seen as a product by rewriting the denominator as left parenthesis g left parenthesis x right parenthesis right parenthesis to the power of negative 1 end exponent

Worked example

Differentiate size 16px space size 16px f begin mathsize 16px style stretchy left parenthesis x stretchy right parenthesis end style size 16px equals fraction numerator size 16px cos size 16px space size 16px 2 size 16px x over denominator size 16px 3 size 16px x size 16px plus size 16px 2 end fraction blank with respect to space x.

5-2-2-ib-sl-aa-only-quotient-we-soltn

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Lucy

Author: Lucy

Expertise: Head of STEM

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels. Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all. Lucy has created revision content for a variety of domestic and international Exam Boards including Edexcel, AQA, OCR, CIE and IB.