CC3 Quotient Rule

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From users who are members of Skills Audits for Maths and Stats :

Will Morgan	said	Ready to use	1 year, 2 months ago
Mash Sheffield	said	Needs to be tested	1 year, 4 months ago

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Will Morgan 1 year ago

Published this.

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Will Morgan 1 year, 2 months ago

Gave some feedback: Ready to use

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Mash Sheffield 1 year, 4 months ago

Gave some feedback: Needs to be tested

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Mash Sheffield 1 year, 4 months ago

Created this as a copy of Differentiation: Quotient Rule 1.

Name	Status	Author	Last Modified
Differentiation: Quotient Rule 1	Ready to use	Ben McGovern	26/09/2022 12:06
Differentiation: Quotient Rule 2	Ready to use	Ben McGovern	26/09/2022 12:06
Differentiation: Quotient Rule 3	Ready to use	Ben McGovern	26/09/2022 12:07
Differentiation: Quotient Rule 4	Ready to use	Ben McGovern	26/09/2022 12:07
Differentiation: Quotient Rule 5	Ready to use	Ben McGovern	26/09/2022 12:08
Differentiation: Quotient Rule 6	Ready to use	Ben McGovern	26/09/2022 12:08
Differentiation: Quotient Rule 7	Ready to use	Ben McGovern	26/09/2022 12:08
CC3 Quotient Rule	Ready to use	Mash Sheffield	17/04/2025 09:02

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Question statement

Find the derivative of

$\simplify{y={a}x^{n}/({b}x+{c})}.$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
b	integer	2
c	integer	-6
n	integer	4
simp	number	18

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Generated value: `integer`

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simp

Statement
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"Unnamed part" → "Gap 0" - Correct answer

Parts

Gap-fill
1. Gap Gap 0 Mathematical expression
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$\dfrac{dy}{dx}=$ Gap 0

Advice

If we have a function of the form $y=\tfrac{u(x)}{v(x)}$ , to calculate its derivative we need to use the quotient rule:

$\dfrac{dy}{dx} = \dfrac{v(x) \times \frac{du}{dx} - u(x) \times\frac{dv}{dx}}{[v(x)]^2}\,.$

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$ ;
Calculate their derivatives $\tfrac{du}{dx}$ and $\tfrac{dv}{dx}$ ;
Substitute these into the formula for the quotient rule to obtain an expression for $\tfrac{dy}{dx}$ ;
Simplify $\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$ .

$\simplify{y={a}x^{n}/({b}x+{c})},$

let

$u(x) = \simplify{{a}x^{n}} \quad \text{and} \quad v(x)=\simplify{{b}x+{c}}.$

Next, we need to find the derivatives, $\tfrac{du}{dx}$ and $\tfrac{dv}{dx}$ :

$\dfrac{du}{dx} = \simplify{{a*n}x^{n-1}}\quad \text{and} \quad\dfrac{dv}{dx}=\simplify{{b}}.$

Substituting these results into the quotient rule formula we can obtain an expression for $\tfrac{dy}{dx}$ :

$\begin{split} \dfrac{dy}{dx} &\,= \dfrac{v(x) \times \frac{du}{dx} - u(x) \times\frac{dv}{dx}}{[v(x)]^2} \\ \\&\,=\dfrac{(\simplify{{b}x+{c}}) \times\simplify{{a*n}x^{n-1}} - \simplify{{a}x^{n}} \times \simplify{{b}}}{\simplify{({b}x+{c})^2}}. \end{split}$

Simplifying,

$\begin{split} \dfrac{dy}{dx} &\,=\dfrac{(\simplify{{b}x+{c}})\simplify{{a*n}x^{n-1}} - \simplify{{b*a}x^{n}}}{\simplify{({b}x+{c})^2}} \\ \\&\,=\dfrac{\simplify[all,!cancelTerms]{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\simplify{({b}x+{c})^2}}\\ \\ &\,=\dfrac{\simplify{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\simplify{({b}x+{c})^2}} \\ \\ &\,=\dfrac{\simplify{{simp}x^{n-1}({(b*a*n-b*a)/simp}x+{c*a*n/simp})}}{\simplify{({b}x+{c})^2}} \end{split}$

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Will Morgan 1 year ago

Will Morgan 1 year, 2 months ago

Mash Sheffield 1 year, 4 months ago

Mash Sheffield 1 year, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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CC3 Quotient Rule

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Will Morgan 1 year ago

Will Morgan 1 year, 2 months ago

Mash Sheffield 1 year, 4 months ago

Mash Sheffield 1 year, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`