Chain rule e^()

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Johnny Yi 6 years, 3 months ago

Created this as a copy of Luis's copy of Chain rule - exponential of polynomial,.

Name	Status	Author	Last Modified
Chain rule - exponential of polynomial,	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
cormac's copy of Chain rule - exponential of polynomial,	draft	cormac breen	15/02/2017 11:39
Harry's copy of cormac's copy of Chain rule - exponential of polynomial,	draft	Harry Flynn	08/05/2018 12:30
Harry's copy of Chain rule - exponential of polynomial,	draft	Harry Flynn	08/05/2018 12:41
Luis's copy of Chain rule - exponential of polynomial,	draft	Luis Hernandez	30/11/2018 19:11
Chain rule e^()	draft	Johnny Yi	19/12/2018 01:07
Musa's copy of 3 Chain rule - exponential of polynomial,	draft	Musa Mammadov	27/06/2023 01:37

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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	4
c	integer	3
b	integer	-4
s2	integer	1
s1	integer	-1
m	integer	3

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Generated value: `integer`

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

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

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

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Advice

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

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

Hence on substituting into the chain rule above we get:

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

$u$ by

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

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Johnny Yi 6 years, 3 months ago

Error in variable testing condition

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Generated value: `integer`

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Chain rule e^()

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Johnny Yi 6 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`