Chain rule - trig and quadratic - Numbas at mathcentre.ac.uk

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From users who are members of Content created by Newcastle University :

Christian Lawson-Perfect

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4 years, 5 months ago

From users who are not members of Content created by Newcastle University :

francisco Glezortiz

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Christian Lawson-Perfect 4 years, 5 months ago

Saved a checkpoint:

I've replaced $dx$ with $\mathrm{d}x$ .

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Christian Lawson-Perfect 4 years, 5 months ago

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francisco Glezortiz 4 years, 6 months ago

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

Created this.

Name	Status	Author	Last Modified
Chain rule - trig and quadratic	Ready to use	Newcastle University Mathematics and Statistics	04/07/2021 16:13
Harry's copy of Chain rule - trig and quadratic,	draft	Harry Flynn	08/05/2018 12:41
Luis's copy of Chain rule - trig and quadratic,	draft	Luis Hernandez	30/11/2018 19:14

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	7
s2	integer	1
c	integer	5
b	integer	6
s1	integer	1

Automatically regenerate variables?

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random(2..9)

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Generated value: `integer`

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

Parts

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1. Gap Mathematical expression
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1. Step Information only

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

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

$\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$

$\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}=$ Unnamed gap

Advice

$\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$

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 $u=\simplify[std]{e^({a}x) +{b}x^2+{c}}$ and we have $f(u)=\simplify[std]{cos(u)}$ .

This gives

$\begin{align} \frac{\mathrm{d}u}{\mathrm{d}x} &= \simplify[std]{{a}e^({a}x) +{2*b}x} \\[1em] \frac{\mathrm{d}f(u)}{\mathrm{d}u} &= \simplify[std]{-sin(u)} \end{align}$

Hence on substituting into the chain rule above we get:

$\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= \simplify[std]{({a}e^({a}x) +{2*b}x) * (-sin(u))} \\ &= \simplify[std]{-({a}e^({a}x) +{2*b}x)*sin(e^({a}x) +{b}x^2+{c})} \end{align}$

on replacing $u$ by $\simplify[std]{e^({a}x) +{b}x^2+{c}}$ .

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Chain rule - trig and quadratic

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Christian Lawson-Perfect 4 years, 5 months ago

Christian Lawson-Perfect 4 years, 5 months ago

francisco Glezortiz 4 years, 6 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Generated value: `integer`

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Chain rule - trig and quadratic

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 4 years, 5 months ago

Christian Lawson-Perfect 4 years, 5 months ago

francisco Glezortiz 4 years, 6 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`