smc-746-30-Chain Rule

Calculus, MET1 2017 VCAA 1b

Let `g(x) = (2-x^3)^3`.

Evaluate `g^{\prime}(1)`. (2 marks)

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`g^{\prime}(1) = -9`

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`text(Using Chain Rule:)`

`g^{\prime}(x)`	`= 3 (2-x^3)^2 (-3x^2)`
	`= -9x^2 (2-x^3)^2`
`:. g^{\prime}(1)`	`= -9 (1) (2-1)^2`
	`= -9`

Calculus, MET1 2019 VCAA 1ai

Let `f: (1/3, oo) -> R,\ \ f(x) = 1/(3x - 1)`.

Find `f prime(x)`. (1 mark)

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`(-3)/(3x – 1)^2`

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	`f(x)`	`= (3x – 1)^(-1)`
	`f′(x)`	`= -1 xx 3xx (3x – 1)^(-2)`
		`= (-3)/(3x – 1)^2`

Calculus, MET1 2018 VCAA 1a

If `y = (−3x^3 + x^2 - 64)^3`, find `(dy)/(dx)`. (1 mark)

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`3(−3x^3 + x^2 – 64)^2 · (−9x^2 + 2x)`

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Mean mark 58%!
MARKER’S COMMENT: Poor use use brackets was notable.

`y`	`= (−3x^3 + x^2 – 64)^3`
`(dy)/(dx)`	`= 3(−3x^3 + x^2 – 64)^2 · (−9x^2 + 2x)`

Calculus, MET1 2014 VCAA 1b

If `f(x) = sqrt(x^2 + 3)`, find `f^{′}(1)`. (3 marks)

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`1/2`

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`text(Using Chain Rule:)`

`f^{′}(x)`	`= 1/2(x^2 + 3)^(-1/2) xx 2x`
`f^{′}(x)`	`= x/(sqrt(x^2 + 3))`
`:. f^{′}(1)`	`=1/(sqrt(1+3))`
	`=1/2`

Calculus, MET2 2009 VCAA 15 MC

For `y = sqrt (1 - f(x)),\ \ (dy)/(dx)` is equal to

A. `(2 f prime (x))/(sqrt(1 - f(x))`

B. `(-1)/(2 sqrt (1 - f prime (x)))`

C. `1/2 sqrt (1 - f prime (x))`

D. `3/(2(1 - f prime(x)))`

E. `(-f prime (x))/(2 sqrt (1 - f (x)))`

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`E`

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`text(Using Chain Rule,)`

`dy/dx`	`= – f′(x) xx 1/2 xx (1-f(x))^(- 1/2)`
	`=(- f′(x))/(2 sqrt (1 – f (x)))`

`=> E`

Calculus, MET1 2008 VCAA 1a

Let `y = (3x^2 - 5x)^5`. Find `(dy)/(dx)`. (2 marks)

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`5(3x^2 – 5x)^4(6x – 5)`

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`text(Using Chain Rule:)`

`(dy)/(dx) = 5(3x^2 – 5x)^4(6x – 5)`

Calculus, MET1 2011 VCAA 1a

Differentiate `sqrt (4 - x)` with respect to `x.` (1 mark)

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`(dy)/(dx) = (– 1)/(2 sqrt (4 – x))`

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`text(Using the Chain Rule:)`

`text(Let)\ \ y`	`= (4 – x)^(1/2)`
`:. (dy)/(dx)`	`= 1/2 (4 – x)^(-1/2) (- 1)`
	`= (– 1)/(2 sqrt (4 – x))`

Calculus, MET1 2012 VCAA 1a

If `y = (x^2 - 5x)^4`, find `(dy)/(dx).` (1 mark)

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`4(x^2 – 5x)^3 (2x – 5)`

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`text(Using Chain Rule:)`

`(dy)/(dx) = 4 (x^2 – 5x)^3 (2x – 5)`

Calculus, MET1 2015 VCAA 1a

Let `y = (5x + 1)^7`.

Find `(dy)/(dx)`. (1 mark)

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`35(5x + 1)^6`

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`text(Using Chain Rule:)`

`(dy)/(dx)`	`= 7(5x + 1)^6 xx 5`
	`= 35(5x + 1)^6`