Principles of Calculus · SHS2 Additional…

Lesson notes

Learning Objectives

Indicator: 2.3.1.LI.2 - Apply the product and quotient rules to differentiate functions.

By the end of the lesson, learners can:

State the product rule formula and the quotient rule formula in their own words.
Apply the product rule to differentiate functions that are products of two differentiable functions, including those involving roots and powers.
Apply the quotient rule to differentiate functions that are quotients of two differentiable functions.
Recognise when it is appropriate to use the product rule, the quotient rule, or a simpler method (such as expanding first) for a given function.
Use the product and quotient rules to solve word problems involving rates of change in practical situations.

Sign in with your phone number to read the full note and download the GES plan - free.

Sign in with phone number

Curriculum details

Strand: Calculus (Strand 3)
Sub-strand: Principles of Calculus (3.1)
Content standard: 2.3.1.CS.1 - Determine the appropriate rule to use in finding the derivative of a function and relations. 2.3.1.LO.1 Determine the appropriate rule and use it to find the derivative of a function. 2.3.1.LO.2 Estimate the area under a curve using the trapezoid rule.
Indicator: 2.3.1.LI.2 - Apply the product and quotient rules to differentiate functions.
Suggested placement: Semester 2, Week 7 (Week 27 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
Source document note: The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:

exemplars - p.380: adjacent duplicated CambriaMath characters were collapsed, but this PDF also maps some equation glyphs to the wrong letter or operator; verify every mathematical expression in this row against the rendered page
exemplars - p.380: a stacked fraction, matrix, vector or other two-dimensional construct is flattened by the text layer; all visible parts require comparison with the rendered page
Curriculum reference: NaCCA curriculum document, p. 380

Exemplars (from the NaCCA curriculum)

Talk for Learning, Think-pair-share, Experiential Learning, and Group Work /Collaborative Learning.
Activity 1: Product and Quotient Rule In mixed-ability groups, learners explore the product and quotient rules to differentiate functions, including composite functions.
Example: Use an appropriate rule of differentiation to find the derivative of the following functions; !
- 𝑓(𝑥) = (0*")
- 𝑓(𝑥) = 4𝑥 − 7𝑥 )
- 𝑓(𝑥) = (2𝑥 + 5𝑥 . )(𝑥 − 2) ()0$.0 ( )(0*))
- 𝑓(𝑥) = (0*") 𝑓(𝑥) = (4𝑥 − 7𝑥 ) )(3𝑥 − 9) !(
-
- 𝑓(𝑥) = (3𝑥 − 4√𝑥) !(
- 𝑓(𝑥) = (3𝑥 − 9) !(
- 𝑓(𝑥) = 𝑠 (2𝑥)
- 𝑓(𝑥) = 𝑠 (2𝑥) 𝑓(𝑥) = 𝑠 (2𝑥 − 𝑥 ) )
-
- 𝑓(𝑥) = 𝑐 (𝑥 ) + 3𝑥) ;PO;PO (0) $;PO ()0)
- 𝑓(𝑥) = (0*")
- 𝑓(𝑥) = 𝑠 ) (𝑥)
- 𝑓(𝑥) = 𝑐 ) (𝑥) + √1 − 𝑥 𝑓(𝑥) = 𝑐 ) (𝑥) + w𝑠 (1 − 𝑥)
-
Activity 2: Learners to Think-pair-share with another learner how to use specific rules on a specific function. E.g.
- 𝑓(𝑥) = 𝑐 ) (𝑥) + √1 − 𝑥
- 𝑓(𝑥) = 𝑐 ) (𝑥) + w𝑠 (1 − 𝑥) !(
- 𝑓(𝑥) = l3𝑥 − 4√𝑥n
- 𝑓(𝑥) = (3𝑥 − 9) !(
Example: Find the derivative of the following: i. 𝑓(𝑥) = 5√3𝑥 ) + 𝑥
Solution: $ 𝑓(𝑥) = 5(3𝑥 ) + 𝑥) ! .(/0$!) ! 𝑓 - (𝑥) = ∙ ) √"0 $! ! .(/0$!) = )√"0 ! $!
! ii. 𝑓(𝑥) = 𝑥 . + 0
Solution: 𝑓(𝑥) = 𝑥 . + 𝑥 *! 𝑓 - (𝑥) = 5𝑥 , − 𝑥 *) = 5𝑥 , − 0 ! !
iii. 𝑦 = 2√𝑥 Solution: $ 𝑦 = 2(𝑥) ! $ <A = 2 ∙ ) (𝑥) * ! ! <0 ! = √0
iv. 𝑦 = (√−2𝑥 + 1)(𝑥 ) + 4) Solution: $ 𝑦 = (−2𝑥 + 1) ! (𝑥 ) + 4) $ $ <A = (−2𝑥 + 1) ! (2𝑥) + (𝑥 ) + 4) ) (−2)(−2𝑥 + 1) * ! ! <0 (0 ! $,) = 2𝑥w(−2𝑥 + 1) − u(*)0$!) )0(*)0$!)*(0 ! $,) = u(*)0$!)
- .0 ! $)0*,) = u(*)0$!)
Teaching and Learning Resources:
- Reading resource
- Colour pens
- Notebook
- Graph sheets
- Mathematical sets
- Technological tools
Assessment (2.3.1.AS.2). The document marks these depth-of-knowledge levels for this indicator: Level 4 Extended critical thinking and reasoning.