Andrew's copy of CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule)

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Andrew Dunbar

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Andrew Dunbar 8 years, 7 months ago

Created this as a copy of CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule).

Name	Status	Author	Last Modified
Differentiation 11 - Product Rule (Basic)	Ready to use	Katie Lester	01/06/2016 09:47
Differentiation 12 - Product Rule (with Chain Rule)	Ready to use	Katie Lester	01/06/2016 09:47
Jinhua's copy of Differentiation 12 - Product Rule (with Chain Rule)	Ready to use	Jinhua Mathias	27/08/2024 16:37
Jinhua's copy of Differentiation 11 - Product Rule (Basic)	Ready to use	Jinhua Mathias	27/08/2024 16:37
Differentiation 11 - Product Rule (Basic)	Ready to use	Jinhua Mathias	28/08/2024 12:15
CF Maths Portfolio - Differentiation 11 - Product Rule (Basic)	Ready to use	Jinhua Mathias	28/08/2024 12:15
CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule)	Ready to use	Jinhua Mathias	28/08/2024 12:15
CF Maths In class test two mock paper question 6 (i)	Ready to use	Jinhua Mathias	28/08/2024 12:15
Vicky's copy of Differentiation 11 - Product Rule (Basic)	draft	Vicky Hall	01/07/2016 20:47
Vicky's copy of Differentiation 11 - Product Rule (Basic)	draft	Vicky Hall	01/07/2016 20:55
Andrew's copy of CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule)	draft	Andrew Dunbar	10/09/2016 22:51
CINCO UNO Derivada producto	draft	david arcila	08/09/2021 00:37
CINCO DOS Derivada producto	draft	Marlon Arcila	08/09/2021 00:38
Product Rule	draft	cormac breen	31/01/2017 13:25
cormac's copy of Product Rule	draft	cormac breen	17/04/2020 19:43
cormac's copy of cormac's copy of Product Rule	draft	cormac breen	07/03/2017 14:16
Milena's copy of cormac's copy of cormac's copy of Product Rule	draft	Milena Venkova	08/03/2018 12:26
Milena's copy of Vicky's copy of Differentiation 11 - Product Rule (Basic)	draft	Milena Venkova	27/02/2018 14:56
Product Rule	draft	Milena Venkova	06/03/2018 14:43
Product Rule 2	draft	Milena Venkova	08/03/2018 15:36
Milena's copy of Product Rule (Basic)	draft	Milena Venkova	08/03/2018 15:45
Product Rule (Basic)3	draft	Milena Venkova	13/03/2018 14:34
Differentiation 12 - Product Rule (with Chain Rule)	draft	Xiaodan Leng	11/07/2019 05:33
Differentiation 11 - Product Rule (Basic)	draft	Xiaodan Leng	11/07/2019 06:56
Differentiation 12 - Product Rule (with Chain Rule)	draft	Xiaodan Leng	11/07/2019 06:56
Musa's copy of 3 Differentiation - Product Rule (Basic)	draft	Musa Mammadov	27/06/2023 01:37

There are 28 other versions that do you not have access to.

Question statement

Differentiate the following expressions with respect to $x$ using the both product rule and the chain rule.

Simplify your answers as much as possible.

Give any introductory information the student needs.

			Name	Type	Generated Value

			c	list	List of 13 items
			p	list	List of 10 items

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Data type

Value

xxxxxxxxxx
 
repeat(random(-9..9 except 0),13)

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `list`

[ -1, -8, 3, -2, 6, -3, -2, -3, 7, 5, 7, -7, 5 ]

→ Used by:

"Part a)" - prompt
"Part a)" - Correct answer
"Part b)" - prompt
"Part b)" - Correct answer

Advice

Patience and careful substitution is all that is necessary to differentiate these types of functions.

Take the function: $y=4x^2(5x^3+2)^4$

First label the two separate parts which would be differentiated with the product rule, namely:

$u=4x^2\;\;\;\text{and}\;\;\;v=(5x^3+2)^4$

We also know that, by the product rule:

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$

Looking at the substitution for $u$ , the derivate can be easily found with the power rule:

$\frac{du}{dx}=(4\times2)x^{2-1}=8x^1=8x$

So far, from the product rule, we have:

$\frac{dy}{dx}=(4x^2)\frac{dv}{dx}+(5x^3+2)^4(8x)$

Looking at the expression for $v=(5x^3+2)^4$ we can see that finding $\frac{dv}{dx}$ , however, requires the chain rule.

Our usual subsitution letter $u$ has already been assigned an expression in the product rule part. This time, we can use $t$ .

By the chain rule:

$\frac{dv}{dx}=\frac{dv}{dt}\times\frac{dt}{dx}$

Let $t=5x^3+2$

This means that: $v=t^4\;\;\;\therefore\;\;\;\frac{dv}{dt}=4t^3$

Also, $t=5x^3+2\;\;\;\therefore\;\;\;\frac{dt}{dx}=15x^2$

It follows that: $\frac{dv}{dx}=\frac{dv}{dt}\times\frac{dt}{dx}=4t^3\times15x^2=60x^2t^3$

Subsituting back the original expression for $t$ :

$\frac{dv}{dx}=60x^2(5x^3+2)^3$

We can now input our expression for $\frac{dv}{dx}$ into the original product rule section.

$\frac{dy}{dx}=(4x^2)\frac{dv}{dx}+(5x^3+2)^4(8x)\;\;\;\therefore\;\;\;\frac{dy}{dx}=(4x^2)(60x^2(5x^3+2)^3)+(5x^3+2)^4(8x)$

This is simplified further:

$\frac{dy}{dx}=240x^4(5x^3+2)^3+8x(5x^3+2)^4=8x(5x^3+2)^3\left(30x^3+(5x^3+2)\right)$

Finally:

$\frac{dy}{dx}=8x(5x^3+2)^3(35x^3+2)$

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Andrew's copy of CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule)

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Andrew Dunbar 8 years, 7 months ago

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Andrew's copy of CF Maths Portfolio - Differentiation 12 - Product Rule (with Chain Rule)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Andrew Dunbar 8 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `list`