Product Rule 06

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Michael Proudman

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Needs to be tested

4 years, 10 months ago

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Michael Proudman 4 years, 10 months ago

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Name	Status	Author	Last Modified
Product Rule 06	Needs to be tested	Michael Proudman	07/12/2021 08:32
Product Rule 06 (non scaffold)	Needs to be tested	Michael Proudman	07/12/2021 08:32

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Question statement

We use the PRODUCT RULE when the function that we need to differentiate is actually two functions multiplied together:

If $y=u \times v$ then:

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

Give any introductory information the student needs.

			Name	Type	Generated Value

bCF	integer	5
bP1	integer	4
bCF2	integer	2

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..6)

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: `integer`

→ Used by:

bCF2
bP1

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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3. Gap Gap 2. Mathematical expression
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4. Gap Gap 3. Mathematical expression
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5. Gap Gap 4. Mathematical expression
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6. Gap Gap 5. Mathematical expression
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7. Gap Gap 6. Mathematical expression
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8. Gap Gap 7. Mathematical expression
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9. Gap Gap 8. Mathematical expression
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Differentiate $y=\var{bCF}x^{\var{bP1}} \cos{(\var{bCF2}x)}$

First identify the two functions $u$ and $v$ :

$u=$ Gap 0. $v=$ Gap 1.

Now differentiate each one:

$\large \frac{du}{dx}=$ Gap 2. $\large\frac{dv}{dx}=$ Gap 3.

Substitute each component into the formula in the correct place:

$\large \frac{dy}{dx}=$ Gap 4. $\large \times$ Gap 5. $\large +$ Gap 6. $\large \times$ Gap 7.

Finally tidy this up to give your final answer:

$\large \frac{dy}{dx}=$ Gap 8.

Advice

We are asked to differentiate:

$y=\var{bCF}x^{\var{bP1}} \cos{(\var{bCF2}x)}$

Recognising that the function to differentiate is the product of two functions, we identify the two functions that are involved.

$u$ is the first function, $v$ is second:

$\large u=\var{bCF}x^{\var{bP1}}$ $\large v=\cos{(\var{bCF2}x)}$

Now, we need to use the approriate techniques to differentiate each of these, for $u$ we can use the Power Rule and for $v$ your Table of Derivatives:

This gives us:

$\large \frac{du}{dx}=\simplify{ {bP1}*{bCF}*x^({bP1}-1) }$ and $\large \frac{dv}{dx}=- \var{bCF2} \sin{(\var{bCF2}x)}$

We now use the formula:

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

Make the appropriate substitutions into the formula:

$\large \frac{dy}{dx}= \var{bCF}x^{\var{bP1}} \times - \var{bCF2}\sin{(\var{bCF2}x)} + \cos{(\var{bCF2}x)} \times \simplify{ {bP1}*{bCF}*x^({bP1}-1) }$

Finally, we need to use our basic algebra to simplify this as much as possible. Multiply out any brackets where it would simplify and collect like terms:

$\large \frac{dy}{dx}= \simplify{ - {bCF2}* {bCF} x^{{bP1}}} \sin{(\var{bCF2}x)} + \simplify{ {bP1}*{bCF}*x^({bP1}-1) } \cos{(\var{bCF2}x)}$

Give a worked solution to the whole question.

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This question is used in the following exam:

Product Rule (Instructional) by Michael Proudman in .Differential Calculus.

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Checkpoint description

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

We use the PRODUCT RULE when the function that we need to differentiate is actually two functions multiplied together:

If $y=u \times v$ then:

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

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Differentiate $y=\var{bCF}x^{\var{bP1}} \cos{(\var{bCF2}x)}$

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Product Rule 06

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

Michael Proudman 4 years, 10 months ago

We use the PRODUCT RULE when the function that we need to differentiate is actually two functions multiplied together:

If y=u×vy=u \times v then:

dydx=udvdx+vdudx \frac{dy}{dx}=u \frac{dv}{dx} + v \frac{du}{dx}

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Differentiate y={bCF}x{bP1}cos({bCF2}x) y=\var{bCF}x^{\var{bP1}} \cos{(\var{bCF2}x)}

If $y=u \times v$ then:

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

Generated value: `integer`

Differentiate $y=\var{bCF}x^{\var{bP1}} \cos{(\var{bCF2}x)}$