Chain Rule 02 (non-scaffolded)

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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, 10 months ago

Created this as a copy of Chain Rule 02.

Name	Status	Author	Last Modified
Chain Rule 02	Needs to be tested	Michael Proudman	07/12/2021 08:32
Chain Rule 02 (non-scaffolded)	Needs to be tested	Michael Proudman	07/12/2021 08:32

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Question statement

We use the CHAIN RULE (also called the FUNCTION OF A FUNCTION RULE) when the function that we need to differentiate is actually one function "nested" inside another.

If $y=f(g(x))$ to find $\frac{dy}{dx}$ , we need to do two things::

Substitute $u=g(x)$ so that $y=f(u)$
Use the Chain Rule:

$\large \frac{dy}{dx}=\frac{du}{dx} \times \frac{dy}{du}$

Now it is time to get out your paper and pencil and try similar questions without the help. Carry out the same steps, lay it out the same way but you only need to input the final answer.

Give any introductory information the student needs.

			Name	Type	Generated Value

			bP	integer	5
			bP2	integer	4

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

Part b) x power

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`

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Mathematical expression
  Add

Gap-fill

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

Differentiate $y=\cos{(x^{\var{bP}})}$

$\large \frac{dy}{dx}=$ Unnamed gap

Advice

We are asked to differentiate:

$y=\cos{(x^{\var{bP}})}$

Recognising that this is a "function of another function", we need to identify the "innermost" of the two functions that are involved and substitute $u$

Let $u=x^{\var{bP}}$ then $y=cos(u)$

Now, we need to use the approriate techniques to differentiate each of these, for the first we need the Power Rule, and for the second you can use your Table of Derivatives.

Applying this method gives us:

$\large \frac{du}{dx}=\simplify{{bP}x^{bP2}}$ and $\large \frac{dy}{du}= -sin(u)$

We now use the Chain Rule formula:

$\large \large \frac{dy}{dx}=\frac{du}{dx} \times \frac{dy}{du}$

Make the appropriate substitutions into the formula:

$\large \frac{dy}{dx}= \simplify{{bP}x^{bP2}} \times -sin(u)$

Which simplifies to:

$\large \frac{dy}{dx}=-\simplify{ {bP}x^{bP2}}sin(u)$

Now, finally, we must remember that $u$ was a variable that we introduced and was not part of the original problem.

Replace $u$ from our original substitution to give the final answer:

$\large \frac{dy}{dx}=-\simplify{{bP}x^{bP2}} sin(x^{\var{bP}})$

Give a worked solution to the whole question.

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This question is used in the following exam:

Chain 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 CHAIN RULE (also called the FUNCTION OF A FUNCTION RULE) when the function that we need to differentiate is actually one function "nested" inside another.

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Differentiate $y=\cos{(x^{\var{bP}})}$

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Error

Chain Rule 02 (non-scaffolded)

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 CHAIN RULE (also called the FUNCTION OF A FUNCTION RULE) when the function that we need to differentiate is actually one function "nested" inside another.

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Differentiate y=cos(x{bP}) y=\cos{(x^{\var{bP}})}

Generated value: `integer`

Differentiate $y=\cos{(x^{\var{bP}})}$