Quotient rule - differentiate quotient of trig functions

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Name	Status	Author	Last Modified
Quotient rule - differentiate quotient of trig functions	Ready to use	Newcastle University Mathematics and Statistics	24/11/2020 13:21
Luis's copy of Quotient rule - differentiate quotient of trig functions	draft	Luis Hernandez	30/11/2018 17:42
Simon's copy of Quotient rule - differentiate quotient of trig functions	draft	Simon Thomas	06/11/2019 09:29
differentiate quotient of trig functions	draft	Xiaodan Leng	10/07/2019 21:16
Quotient rule - differentiate quotient of trig functions	draft	Xiaodan Leng	10/07/2019 22:43

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Question statement

Differentiate the following functions using the quotient rule.

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			Name	Type	Generated Value

a	integer	4
c	integer	9
b	integer	8
s2	integer	-1
s1	integer	1
m	integer	2
n	integer	-3
res	integer	168
n1	integer	-3
c1	integer	8

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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

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

a)

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:

$\begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= \simplify[std]{({a}cos({a}x)({b}sin({a}x)+{c}cos({a}x))-sin({a}x)({a*b}cos({a}x)+{-a*c}sin({a}x)))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({a*b} cos({a}x) sin({a}x)+{a*c} cos({a}x)^2-{a*b} sin({a}x)cos({a}x)+{a*c}sin({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({a*c}cos({a}x)^2+{a*c}sin({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({a*c}(cos({a}x)^2+sin({a}x)^2))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({a*c})/({b}sin({a}x)+{c}cos({a}x))^2} \end{align}$

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

b)

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

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

Hence on substituting into the quotient rule above we get:

$\begin{align} \frac{\mathrm{d}g}{\mathrm{d}x} &= \simplify[std]{({-a}sin({a}x)({b}sin({a}x)+{c}cos({a}x))-cos({a}x)({a*b}cos({a}x)+{-a*c}sin({a}x)))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({-a*b}sin({a}x)^2-{a*c} sin({a}x)cos({a}x)-{a*b}cos({a}x)^2+{a*c}sin({a})cos({a}x))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({-a*b}sin({a}x)^2-{a*b}cos({a}x)^2)/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({-a*b}(sin({a}x)^2+cos({a}x)^2))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{({-a*b})/({b}sin({a}x)+{c}cos({a}x))^2} \end{align}$

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

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

Hence

$\begin{align} \frac{\mathrm{d}h}{\mathrm{d}x} &= \simplify[std]{{m}*Diff(f,x,1)+{n}*Diff(f,x,1)} \\[0.5em] &= \simplify[std]{{m}*({a*c}/({b}sin({a}x)+{c}cos({a}x))^2)+{n}({-a*b}/({b}sin({a}x)+{c}cos({a}x))^2)} \\[0.5em] &= \simplify[std]{(({m}*{a*c})+({n}*{-a*b}))/({b}sin({a}x)+{c}cos({a}x))^2} \\[0.5em] &= \simplify[std]{{res}/({b}sin({a}x)+{c}cos({a}x))^2} \end{align}$

Hence $c=\var{res}$ .

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lynette O'Callaghan 4 years, 5 months ago

Christian Lawson-Perfect 4 years, 6 months ago

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Quotient rule - differentiate quotient of trig functions

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Taxonomy: mathcentre

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lynette O'Callaghan 4 years, 5 months ago

Christian Lawson-Perfect 4 years, 6 months ago

Christian Lawson-Perfect 4 years, 6 months ago

Bill Foster 6 years, 3 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Warning

Generated value: integer

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b)

Generated value: `integer`