6G - Chain, product and quotient rules:
The chain product and quotient rules are used to differentiate more complex functions.
The chain rule:
■ A chain rule is used to differentiate composite functions; that is, when a function has an "inner" function and an "outer function". For the composite function \(y=f(g(x))\) the derivative is given by \(\frac{dy}{dx}=g'(x)\times f'(g(x))\).
■ On the VCAA formula sheet, where \(u\) is the "inner function": \[\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}\]
■ On the VCAA formula sheet, where \(u\) is the "inner function": \[\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}\]
6G - Example 1: Differentiation using the chain rule
If \(y=e^{x^2+2}\), calculate \(\frac{dy}{dx}\).
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6G - Example 1: Video solution
6G - Example 1: Practice
Question 1: ABC Question 2: ABC 6G - Example 1: Solutions
Question 1: ABC Question 2: ABC |
6G - Example 2: Differentiation using the chain rule
If \(y=6\sqrt{x^2+5x}\), calculate \(\frac{dy}{dx}\).
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6G - Example 2: Video solution
6G - Example 2: Practice
Question 1: If \(y=cos(e^{2x})\), calculate \(\frac{dy}{dx}\) Question 2: ABC 6G - Example 2: Solutions
Question 1: \[\frac{dy}{dx}=-2e^{2x}sin(e^{2x})\] Question 2: ABC |
6G - Example 3: Differentiation using the chain rule
Simplify the following expression and hence calculate the derivative, \(g'(x)\). \[g(x)=\frac{\sqrt{f(x)}}{f(x)}\]
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6G - Example 3: Video solution
6G - Example 3: Practice
Question 1: ABC Question 2: ABC 6G - Example 3: Solutions
Question 1: ABC Question 2: ABC |
6G - Example 4: Differentiation using the chain rule (VCAA, 2007)
[VCAA, 2007 Exam 1 Question 2]
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6G - Example 4: Video solution
6G - Example 4: Practice
Question 1: ABC Question 2: ABC 6G - Example 4: Solutions
Question 1: ABC Question 2: ABC |
The product rule:
■ A product rule is used to differentiate functions of the form \(f(x) \times g(x)\); that is, when an expression contains two function multiplied together. For the function \(y=f(x)g(x)\) the derivative is given by \(\frac{dy}{dx}=f(x)g'(x)+g(x)f'(x)\).
■ On the VCAA formula sheet, where \(u\) and \(v\) are functions: \[\frac{d}{dx}(uv)=u\frac{dv}{dx} + v\frac{du}{dx}\]
■ On the VCAA formula sheet, where \(u\) and \(v\) are functions: \[\frac{d}{dx}(uv)=u\frac{dv}{dx} + v\frac{du}{dx}\]
6G - Example 5: Differentiation using the product rule
If \(f(x)=x^2e^{-4x}\), calculate the derivative, \(f'(x)\).
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6G - Example 5: Video solution
6G - Example 5: Practice
Question 1: ABC Question 2: ABC 6G - Example 5: Solutions
Question 1: ABC Question 2: ABC |
6G - Example 6: Differentiation using the product rule
If \(g(x)=xsin(x)\), evaluate \(g'(\pi)\).
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6G - Example 6: Video solution
6G - Example 6: Practice
Question 1: ABC Question 2: ABC 6G - Example 6: Solutions
Question 1: ABC Question 2: ABC |
The quotient rule:
■ A quotient rule is used to differentiate functions of the form \(\frac{f(x)}{g(x)}\); that is, when an expression contains the division of a function by another function.
■ On the VCAA formula sheet, where \(u\) and \(v\) are functions: \[\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\]
■ On the VCAA formula sheet, where \(u\) and \(v\) are functions: \[\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\]
6G - Example 7: Differentiation using the quotient rule
Calculate the derivative, \(\frac{dy}{dx}\) for the following function \[y=\frac{x^2+4x}{3-2x}\]
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6G - Example 7: Video solution
6G - Example 7: Practice
Question 1: ABC Question 2: ABC 6G - Example 7: Solutions
Question 1: ABC Question 2: ABC |
6G - Example 8: Differentiation using the quotient rule
If \(f(x)=\frac{2e^{3x}}{cos(2x)}\), evaluate \(f'(0)\).
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6G - Example 8: Video solution
6G - Example 8: Practice
Question 1: ABC Question 2: ABC 6G - Example 8: Solutions
Question 1: ABC Question 2: ABC |
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Please view Topic Worksheets for any additional questions available at this point in time.
Question 1
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Solutions:
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Chain, product and quotient rules - Worksheet A
Worksheet contains past VCAA Exam 1 questions from 2006 to 2010.
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Chain, product and quotient rules - Worksheet B
Worksheet contains past VCAA Exam 1 questions from 2011 to 2015.
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Chain, product and quotient rules - Worksheet C (Coming Soon)
Worksheet contains past VCAA Exam 1 questions from 2016 to 2018.
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Past VCAA examination questions for differentiation using the chain, product and quotient rules have been organised into Topic Worksheets. Please access these worksheets from the Topic Worksheets Tab above.
Nancy Pi
Nancy Pi: The chain rule... How? When?
Nancy Pi introduces the chain rule and goes through several different examples.
(1) Find \(\frac{dy}{dx}\) if \(y=(3x+1)^7\). How do you know when you need to use the chain rule? (2) Find the derivative of \(h(x)=(x^2+5x-6)^9\). (3) Find \(\frac{dy}{dx}\) if \(y=sin(x^2-3x)\). P.S. Double chain rule?! |
nancy Pi: Video Tutorial
Nancy Pi: Solutions
(1) \(\frac{dy}{dx}=21(3x+1)^6\) (2) \(h'(x)=9(2x+5)(x^2+5x-6)^8\) (3) \(\frac{dy}{dx}=(2x-3)cos(x^2-3x)\) |
Nancy Pi: The chain rule... More chain rule
Nancy Pi goes through several more examples using the chain rule (introduced in previous video).
(4) Find \(\frac{dy}{dx}\) if \(y=e^3x\). (5) Find \(\frac{dy}{dx}\) if \(y=log_{e}(5x)\). (6) Find \(\frac{dy}{dx}\) if \(y=\sqrt{x^2+1}\). (7) Find \(\frac{dy}{dx}\) if \(y=x^3(2x-5)^4\). |
nancy Pi: Video Tutorial
Nancy Pi: Solutions
(4) \(\frac{dy}{dx}=3e^3x\) (5) \(\frac{dy}{dx}=\frac{1}{x}\) (6) \(\frac{dy}{dx}=\frac{x}{\sqrt{x^2+1}}\) (7) \(\frac{dy}{dx}=8x^3(2x-5)^3+3x^2(2x-5)^4\) |
Nancy Pi: Derivatives... How? (Power rule, Product rule, Quotient rule)
Nancy Pi introduces many of the rules for differentiation and goes through several different examples.
The Power Rule: (1) Find \(\frac{d}{dx}(x^4+2x^3-x^2+4x-1)\). (2) Find \(\frac{d}{dx}(x^-2)\). (3) Find \(\frac{d}{dx}(x^{\frac{4}{3}})\). The Product Rule: (4) Find \(\frac{dy}{dx}\) if \(y=(x^2+1)(x^3+x+5\). The Quotient Rule: (5) Find \(f'(x)\) if \(f(x)=\frac{x^2+1}{3x-2}\). |
nancy Pi: Video Tutorial
Nancy Pi: Solutions
(1) \(4x^3+6x^2-2x+4\) (2) \(-2x^{-3}=\frac{-2}{x^3}\) (3) \(\frac{4}{3}x^{\frac{1}{3}}\) (4) \((x^2+1)(3x^2+1)+(x^3+x+5)(2x)\) (5) \(\frac{(3x-2)(2x)-(x^2+1)(3)}{(3x-2)^2}\) |