Maths Support: 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.

$a=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

$b=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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b)

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

$\displaystyle \frac{dg}{dx}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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

$\displaystyle \frac{df}{dx}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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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.

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 $g(x)$

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Input numbers as fractions or integers and not as decimals.

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

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.

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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 $a$.

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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 $b$.

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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 $c$.

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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:

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

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

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

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 $g(x)$

$g(x)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

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.

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This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback.

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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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Maths Support: Quotient rule

Question 1

Question 1

a)

b)

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Question 2

Question 2

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Question 3

Question 3

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Question 4

Question 4

a)

b)

c)

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Question 5

Question 5

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