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	2
c	integer	-8
b	integer	-5
s2	integer	-1
s1	integer	-1
m	integer	4

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

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

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$\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\simplify[std]{ln(a^r)=r*ln(a)}$ .
Hence $\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\simplify[std]{ln({a}x+{b})}$
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 $u=\simplify[std]{{a}x +{b}}$ and we have $f(u)=\simplify[std]{{m}*ln(u)}$ .
This gives

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

Hence on substituting into the chain rule above we get:

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

$u$ by

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

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Bill Foster 55 years, 2 months ago

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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

Checkpoint description

Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`