Product Rule 03

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Michael Proudman

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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

Gave some feedback: Needs to be tested

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Michael Proudman 4 years, 10 months ago

Published this.

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Michael Proudman 4 years, 11 months ago

Created this as a copy of Product Rule 04.

Name	Status	Author	Last Modified
Product Rule 04	Needs to be tested	Michael Proudman	07/12/2021 08:32
Product Rule 03	Needs to be tested	Michael Proudman	07/12/2021 08:32
Product Rule 03 (non scaffold)	Needs to be tested	Michael Proudman	07/12/2021 08:32
Product Rule 04 (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

			a	integer	4
			b	integer	8

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Parts

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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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9. Gap Gap 8. Mathematical expression
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Ask the student a question, and give any hints about how they should answer this part.

Differentiate $y=e^{\var{a}x} \ln{(\var{b}x)}$

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.

Then using:

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

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:

$y=e^{\var{a}x} \ln{(\var{b}x)}$

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:

$u=e^{\var{a}x}$ $v=\ln{(\var{b}x)}$

Now, we need to use the approriate techniques to differentiate each of these, for these functions we need your Table of Derivatives:

This gives us:

$\large \frac{du}{dx}=\var{a}e^{\var{a}x}$ and $\large \frac{dv}{dx}= \frac{1}{x}$

We now use the formula:

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

Make the appropriate substitutions into the formula:

$\large \frac{dy}{dx}= e^{\var{a}x} \times \frac{1}{x} + \ln{(\var{b}x)} \times \var{a}e^{\var{a}x}$

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

$\large \frac{dy}{dx}= \frac{ e^{\var{a}x}}{x} + \var{a}e^{\var{a}x} \ln{(\var{b}x)}$

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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, 11 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 $y=e^{\var{a}x} \ln{(\var{b}x)}$

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Product Rule 03

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, 11 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=e{a}xln({b}x) y=e^{\var{a}x} \ln{(\var{b}x)}

If $y=u \times v$ then:

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

Generated value: `integer`

Differentiate $y=e^{\var{a}x} \ln{(\var{b}x)}$