Quotient rule

Question 1

Let \[\simplify[std]{f(x) = ({a} * x+{b})/({c}x+{d})}\]

a)

Find numbers $a$ and $b$ such that
\[\simplify[std]{f(x) = a + b/({c}x+{d})}\]
Enter a and b as integers or fractions, but not as decimals.

$a=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{3}$

$b=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{3}$

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$\simplify[std]{{a}x+{b}=a*({c}x+{d})+b}$ for suitable numbers $a$ and $b$.

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b)

Differentiate
\[\simplify[std]{g(x) = 1/({c}x+{d})}\]

$\displaystyle \frac{dg}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-\frac{ 3 }{ \left ( 3 x + 2 \right )^{ 2 } }}$

Hence using the first part of the question differentiate \[\simplify[std]{f(x) = ({a} * x+{b})/({c}x+{d})}\]

$\displaystyle \frac{df}{dx}=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-\frac{ 9 }{ \left ( 3 x + 2 \right )^{ 2 } }}$

Input numbers as fractions or integers and not as decimals.

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Advice

a)

We have $\displaystyle \simplify[std]{{a}x+{b}={a}/{c}*({c}x+{d})+{b}-{a}*{d}/{c}={a}/{c}*({c}x+{d})+{-det}/{c}}$
Hence \[\begin{eqnarray*} \simplify[std]{({a} * x+{b})/({c}x+{d})}&=&\simplify[std]{({a}/{c}*({c}x+{d})+{-det}/{c})/({c}x+{d})}\\ &=&\simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})} \end{eqnarray*}\]
Where we have divided out by $\simplify[std]{{c}x+{d}}$ at the last step.

b)

We have \[\frac{dg}{dx} = \simplify[std]{{-c}/({c}x+{d})^2}\]
using standard rules of differentiation.
Since from a), \[f(x) = \simplify[std]{{a}/{c}+({-det}/{c})/({c}x+{d})}\]
we see that
\[\begin{eqnarray*}\frac{df}{dx} &=&\simplify[std,fractionNumbers,!unitPower,!zeroFactor,!zeroTerm,!zeroPower]{(-{c})*(({-det}/{c})/({c}x+{d})^2)}\\ &=&\simplify[std]{{det}/({c}x+{d})^2} \end{eqnarray*}\]

Question 2

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

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

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-56 x^{ 2 } - 28 x - 12}$

Input numbers as fractions or integers and not as decimals.

Click on Show steps for more information. You will not lose any marks by doing so.

The quotient rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\]

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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:

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

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

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

Question 3

Differentiate the following function $f(x)$ using the quotient rule or otherwise.

\[\simplify[std]{f(x) = ({a} * x + {b}) / Sqrt({c} * x + {d})}\]

You are given that \[\simplify[std]{Diff(f,x,1) = g(x) / (2 * ({c} * x + {d}) ^ (3 / 2))}\]

for a polynomial $g(x)$. You have to find $g(x)$.

Input all numbers as fractions or integers.

You can click on Show steps to get help. You will not lose any marks if you do so.

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{30 x - 59}$

The quotient rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u/v,x,1)=(v * Diff(u,x,1) -(u * Diff(v,x,1))) / v ^ 2}\]

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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:

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

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

Hence on substituting into the quotient rule above we get:

\[\simplify[std]{Diff(f,x,1) = ({a} * Sqrt({c} * x + {d}) -(({a} * x + {b}) * Diff(v,x,1))) / ({c} * x + {d}) = ({a} * Sqrt({c} * x + {d}) -(({c} * ({a} * x + {b})) / (2 * Sqrt({c} * x + {d})))) / ({c} * x + {d}) = ({2 * a} * ({c} * x + {d}) -({c} * ({a} * x + {b}))) / (2 * ({c} * x + {d}) ^ (3 / 2)) = ({a * c} * x + {2 * a * d -(c * b)}) / (2 * ({c} * x + {d}) ^ (3 / 2))}\]

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

Question 4

Differentiate the following functions using the quotient rule.

a)

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

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 $a$.

$a=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-10}$

The quotient rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\]

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b)

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

You are given that \[\simplify[std]{Diff(g,x,1) = b / ({b}sin({a}x)+{c}cos({a}x))^2}\]

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

$b=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-40}$

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c)

\[\simplify[std]{h(x) = ({m}sin({a}x)+{n}cos({a}x))/({b}sin({a}x)+{c}cos({a}x))}\]

You are given that \[\simplify[std]{Diff(h,x,1) = c / ({b}sin({a}x)+{c}cos({a}x))^2}\]

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

$c=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-310}$

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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:

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

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

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:

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

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

c)

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 $c=\var{res}$

Question 5

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

\[\simplify[std]{f(x) = ({a} * x^2+{b})/({c}x^2+{d})}\]
You are given that \[\frac{df}{dx}=\simplify[std]{g(x)/({c}x^2+{d})^2}\]
for a polynomial $g(x)$. You are asked to find $g(x)$

$g(x)=\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-98 x}$

Input numbers as fractions or integers and not as decimals.

Click on Show steps for more information. You will not lose any marks by doing so.

You will also find a video in Show steps showing how to apply the quotient rule.

The quotient rule says that if $u$ and $v$ are functions of $x$ then
\[\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\]

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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:

\[\simplify[std]{u = ({a}x^2+{b})}\Rightarrow \simplify[std]{Diff(u,x,1) = {2*a}x}\]

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

Hence on substituting into the quotient rule above we get:

\[\begin{eqnarray*} \frac{df}{dx}&=&\simplify[std]{({2*a}x({c}x^2+{d})-{2*c}x({a}x^2+{b}))/({c}x^2+{d})^2}\\ &=&\simplify[std]{({2*a*c}x^3+{2*a*d}x-{2*c*a}x^3-{2*c*b}x)/({c}x^2+{d})^2}\\ &=&\simplify[std]{({2*det}x)/({c}x^2+{d})^2} \end{eqnarray*}\]
Hence $g(x)=\simplify[std]{{2*det}x}$

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Quotient rule

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