Chain rule - power of a quadratic, - Numbas at mathcentre.ac.uk

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Name	Status	Author	Last Modified
Chain rule - power of a quadratic,	Ready to use	Newcastle University Mathematics and Statistics	05/10/2020 12:27
Adrian's copy of Chain rule - power of a quadratic,	draft	Adrian Jannetta	16/08/2023 14:56
Harry's copy of Chain rule - power of a quadratic,	draft	Harry Flynn	08/05/2018 12:40
Luis's copy of Chain rule - power of a quadratic,	draft	Luis Hernandez	30/11/2018 19:12
JD's copy of Chain rule - power of a quadratic,	draft	JD Ichwan	21/05/2019 05:39

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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	5
c	integer	-8
b	integer	5
s2	integer	-1
s1	integer	1
m	integer	6
n	integer	5

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Generated value: `integer`

Parts

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

$\simplify[std]{f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}$

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

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Advice

$\simplify[std]{f(x) = f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}$
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 $u=\simplify[std]{{a} * x^{m}+{c}*x^2+{b}}$ and we have $f(u)=\simplify[std]{u^{n}}$ .
This gives

$\begin{eqnarray*}\frac{du}{dx} &=& \simplify[std]{{m*a}x ^ {m -1}+{2*c}x}\\ \frac{df(u)}{du} &=& \simplify[std]{{n}u^{n-1}} \end{eqnarray*}$

Hence on substituting into the chain rule above we get:

$\begin{eqnarray*}\frac{df}{dx} &=& \simplify[std]{({m*a}x ^ {m-1}+{2*c}x) * ({n}*u^{n-1})}\\ &=&\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)u^{n-1}}\\ &=& \simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)({a}*x^{m}+{c}*x^2+{b})^{n-1}} \end{eqnarray*}$
on replacing

$u$ by

$\simplify[std]{{a}x^{m}+{c}x^2+{b}}$ .

Give a worked solution to the whole question.

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Chain rule - power of a quadratic,

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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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Chain rule - power of a quadratic,

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

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Comment

Checkpoint description

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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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`