Quotient 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 quotient rule.

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			Name	Type	Generated Value

a	integer	3
c	integer	1
b	integer	-4
d	integer	-9
f	integer	-4
s2	integer	-1
s1	integer	-1
det	integer	-48
c1	integer	1

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$\simplify[std]{f(x) = ({a} * x+{b})/({c}x^2+{d}x+{f})}$
You are given that

$\frac{df}{dx}=\simplify[std]{g(x)/({c}x^2+{d}x+{f})^2}$
for a polynomial

$g(x)$ . You are asked to find

$g(x)$

$g(x)=\;$ Unnamed gap

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The quotient rule says that if $u$ and $v$ are functions of $x$ then

$\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}$

For this example:

$\simplify[std]{u = ({a}x+{b})}\Rightarrow \simplify[std]{Diff(u,x,1) = {a}}$

$\simplify[std]{v = ({c} * x^2+{d}x+{f})} \Rightarrow \simplify[std]{Diff(v,x,1) = {2*c}x+{d}}$

Hence on substituting into the quotient rule above we get:

$\begin{eqnarray*} \frac{df}{dx}&=&\simplify[std]{({a}({c}x^2+{d}x+{f})-({2*c}x+{d})({a}x+{b}))/({c}x^2+{d}x+{f})^2}\\ &=&\simplify[std]{({a*c}x^2+{a*d}x+{a*f}-{2*c*a}x^2-{a*d+2*c*b}x+{d*b})/({c}x^2+{d}x+{f})^2}\\ &=&\simplify[std]{({-c*a}x^2+{-2*b*c}x+{a*f-d*b})/({c}x^2+{d}x+{f})^2} \end{eqnarray*}$
Hence

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

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Bill Foster 55 years, 2 months ago

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Generated value: `integer`

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Quotient 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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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`