Harry's copy of Chain rule - binomial, - Numbas at mathcentre.ac.uk

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Harry Flynn

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Harry Flynn 6 years, 10 months ago

Created this as a copy of Chain rule - binomial.

Name	Status	Author	Last Modified
Chain rule - binomial	Ready to use	Newcastle University Mathematics and Statistics	04/12/2020 23:48
Harry's copy of Chain rule - binomial,	draft	Harry Flynn	08/05/2018 12:40
Luis's copy of Chain rule - binomial,	draft	Luis Hernandez	30/11/2018 19:11
Musa's copy of 3 Chain rule - binomial,	draft	Musa Mammadov	27/06/2023 01:36
Derivative (Function of a Function)	Ready to use	Shaheen Charlwood	27/10/2022 18:51

There are 26 other versions that do you not have access to.

Question statement

Differentiate the following function $f(x)$ using the chain rule.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	4
s1	integer	1
b	integer	5
m	integer	5
n	integer	5

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..9)

JME function reference

Description

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

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Generated value: `integer`

→ Used by:

Advice
"Unnamed part" - prompt
"Unnamed part" → "Unnamed gap" - Correct answer

Parts

Gap-fill
1. Gap Mathematical expression
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1. Step Information only

Gap-fill

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

$\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$

$\displaystyle \frac{df}{dx}=\;$ Unnamed gap

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Advice

$\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
The chain rule says that if $f(x)=g(h(x))$ then

$\simplify[std]{f'(x) = h'(x)g'(h(x))}$
One way to find

$f'(x)$ is to let

$u=h(x)$ then we have

$f(u)=g(u)$ as a function of

$u$ .
Then we use the chain rule in the form:

$\frac{df}{dx} = \frac{du}{dx}\frac{df(u)}{du}$
Once you have worked this out, you replace

$u$ by

$h(x)$ and your answer is now in terms of

$x$ .

For this example, we let $u=\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\simplify[std]{u^{n}}$ .
This gives

$\begin{eqnarray*}\frac{du}{dx} &=& \simplify[std]{{m*a}x ^ {m -1}}\\ \frac{df(u)}{du} &=& \simplify[std]{{n}u^{n-1}} \end{eqnarray*}$

Hence on substituting into the chain rule above we get:

$\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{{m*a}x ^ {m-1} * ({n}*u^{n-1})}\\ &=&\simplify[std]{{m*a*n}x^{m-1}u^{n-1}}\\ &=& \simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}} \end{eqnarray*}$
on replacing

$u$ by

$\simplify[std]{{a}x^{m}+{b}}$ .

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Harry's copy of Chain rule - binomial,

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Taxonomy: mathcentre

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Harry Flynn 6 years, 10 months ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

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Error

Harry's copy of Chain rule - binomial,

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Harry Flynn 6 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`