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 functions using the quotient rule.

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			Name	Type	Generated Value

a	integer	2
c	integer	5
b	integer	9
s2	integer	-1
s1	integer	1
m	integer	8
n	integer	-3
res	integer	134
n1	integer	-3
c1	integer	5

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$\simplify[std]{f(x) = (sin({a}x))/({b}sin({a}x)+{c}cos({a}x))}$

You are given that

$\simplify[std]{Diff(f,x,1) = a / ({b}sin({a}x)+{c}cos({a}x))^2}$

for a number $a$ . You have to find $a$ .

$a=\;$ Unnamed gap

Advice

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:

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

$\simplify[std]{v = {b}sin({a}x)+{c}cos({a}x)} \Rightarrow \simplify[std]{Diff(v,x,1) = {a*b}cos({a}x)+{-a*c}sin({a}x)}$

Hence on substituting into the quotient rule above we get:

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

Hence $a=\var{a*c}$

$\simplify[std]{u = cos({a}x)}\Rightarrow \simplify[std]{Diff(u,x,1) = -{a}sin({a}x)}$

$\simplify[std]{v = {b}sin({a}x)+{c}cos({a}x)} \Rightarrow \simplify[std]{Diff(v,x,1) = {a*b}cos({a}x)+{-a*c}sin({a}x)}$

Hence on substituting into the quotient rule above we get:

$\begin{eqnarray*} \frac{dg}{dx}&=&\simplify[std]{({-a}sin({a}x)({b}sin({a}x)+{c}cos({a}x))-cos({a}x)({a*b}cos({a}x)+{-a*c}sin({a}x)))/({b}sin({a}x)+{c}cos({a}x))^2}\\ &=&\simplify[std]{({-a*b}sin({a}x)^2-{a*c} sin({a}x)cos({a}x)-{a*b}cos({a}x)^2+{a*c}sin({a})cos({a}x))/({b}sin({a}x)+{c}cos({a}x))^2}\\ &=&\simplify[std]{({-a*b}sin({a}x)^2-{a*b}cos({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2}\\ &=&\simplify[std]{({-a*b}(sin({a}x)^2+cos({a}x)^2))/({b}sin({a}x)+{c}cos({a}x))^2}\\ &=&\simplify[std]{({-a*b})/({b}sin({a}x)+{c}cos({a}x))^2} \end{eqnarray*}$

Hence $b=\var{-a*b}$

We have that $h(x)=\simplify[std]{{m}f(x)+{n}g(x)}$
Hence

$\begin{eqnarray*}\frac{dh}{dx} &=& \simplify[std]{{m}*Diff(f,x,1)+{n}*Diff(f,x,1)}\\ &=&\simplify[std]{{m}*({a*c}/({b}sin({a}x)+{c}cos({a}x))^2)+{n}({-a*b}/({b}sin({a}x)+{c}cos({a}x))^2)}\\ &=&\simplify[std]{(({m}*{a*c})+({n}*{-a*b}))/({b}sin({a}x)+{c}cos({a}x))^2}\\ &=&\simplify[std]{{res}/({b}sin({a}x)+{c}cos({a}x))^2} \end{eqnarray*}$

Hence $c=\var{res}$

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This question is used in the following exams:

Bradford: Quotient rule practice by Bill Foster in Bill's workspace.
Maths Support: Quotient rule by Christian Lawson-Perfect in Maths Support Wiki.
mathcentre: Quotient rule practice by Bill Foster in Bill's workspace.
Sheet 2 by Stephen Griffiths in Stephen's workspace.
Basic Calculus by Bill Foster in Bill's workspace.
Bradford: Basic Calculus by Bill Foster in Bill's workspace.
Stanislav's copy of Bradford: Basic Calculus by Stanislav Senveter in Stanislav's workspace.
Quotient rule by Marlon Arcila in Derivadas.
Blathnaid's copy of Quotient rule by Blathnaid Sheridan in Blathnaid's workspace.

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Bill Foster 55 years, 2 months ago

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Quotient rule

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Taxonomy: mathcentre

Taxonomy: Kind of activity

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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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Generated value: `integer`