Jinhua's copy of Differentiation 11 - Product Rule (Basic)

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Katie Lester

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Ready to use

8 years, 10 months ago

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Katie Lester 8 years, 10 months ago

Gave some feedback: Ready to use

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Jinhua Mathias 9 years, 5 months ago

Created this as a copy of Differentiation 11 - Product Rule (Basic).

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 product rule.

Simplify your answers as much as possible.

Give any introductory information the student needs.

			Name	Type	Generated Value

c	list	List of 11 items
p	list	List of 10 items
c1	integer	-8

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
repeat(random(-9..9 except 0),11)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `list`

[ 5, -2, 7, 3, 5, 4, -5, -7, -6, -4, -6 ]

→ Used by:

Advice

These questions use the product rule.

The product rule is defined as

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

when $y=uv$

Worked example using Part a:

$y=\simplify{(x+{c[0]})(x+{c1})}$

This expression is the result of two products; $(\simplify{x+{c[0]}})$ and $(\simplify{x+{c1}})$ .

We can therefore say:

$u=(\simplify{x+{c[0]}})$

and

$v=(\simplify{x+{c1}})$ ,

Hence meaning that $y=uv$ .

We already have what $u$ and $v$ equal, so all we have to do is find what $\frac{du}{dx}$ and $\frac{dv}{dx}$ are, and then substitute everything into the rule.

Differentiating with respect to $x$ , we get:

$\frac{du}{dx}=1$

and

$\frac{dv}{dx}=1$ .

As there are no powers or coefficients of $x$ that are $>1$ , this is a very simple version of the product rule, but knowing how to work out this equation formally will make more difficult looking problems just as simple.

Substituting in all the results we've found, we get:

$\frac{dy}{dx}=1(\simplify{x+{c1}})+1(\simplify{x+{c[0]}})$

We then simplify, collecting all the terms, to get our final answer of:

$\frac{dy}{dx}=\simplify{2x+{c[0]}+{c1}}$

Give a worked solution to the whole question.

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This question is used in the following exam:

Differentiation 2 by Paul Finley in Paul's workspace.

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Jinhua's copy of Differentiation 11 - Product Rule (Basic)

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Katie Lester 8 years, 10 months ago

Jinhua Mathias 9 years, 5 months ago

Error in variable testing condition

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Generated value: `list`

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Mathematical expression

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Jinhua's copy of Differentiation 11 - Product Rule (Basic)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Katie Lester 8 years, 10 months ago

Jinhua Mathias 9 years, 5 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`