Chain rule

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Bill Foster

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Bill Foster 55 years, 2 months ago

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Question statement

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

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			Name	Type	Generated Value

a	integer	6
s1	integer	1
b	integer	5
m	integer	3

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

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

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Advice

$\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}$
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 $f(u)=\simplify[std]{sqrt(u)=u^{1/2}}$ .
This gives

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

Hence on substituting into the chain rule above we get:

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

$u$ by

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

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This question is used in the following exams:

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Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Chain rule

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

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Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: integer

Parts

Gap-fill

Generated value: `integer`