smc-744-50-Quotient Rule

Calculus, MET1 2022 VCAA 1b

Find and simplify the rule of `f^{\prime}(x)`, where `f:R \rightarrow R, f(x)=\frac{\cos (x)}{e^x}`. (2 marks)

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`\frac{-(\sin x+\cos x)}{e^x}`

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Using the quotient rule

`f(x)`	`=\frac{\cos (x)}{e^x}`
`f^{\prime}(x)`	`=\frac{-e^x \sin x-e^x \cos x}{e^{2 x}}`
	`= \frac{-e^x(\sin x+\cos x)}{e^{2 x}}`
	`=\frac{-(\sin x+\cos x)}{e^x}`

Calculus, MET1 2006 ADV 2aii

Differentiate with respect to `x`:

Let `y=sin x/(x + 1)`. Find `dy/dx `. (2 marks)

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`dy/dx = {cos x (x + 1)-sin x} / (x + 1)^2`

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`y = sinx/(x + 1)`

`d/dx (u/v) = (u^{\prime} v-uv^{\prime})/v^2`

`u`	`= sin x`	`v`	`= x + 1`
`u^{\prime}`	`= cos x`	`\ \ \ v^{\prime}`	`= 1`

`:.dy/dx = {cos x (x + 1)-sin x} / (x + 1)^2`

Calculus, MET1 2019 VCAA 1b

Let `g: R text(\ {−1}) -> R,\ \ g(x) = (sin(pi x))/(x + 1)`.

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

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`-pi/2`

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	`u = sin(pi x)`	`v = x + 1`
	`u^{prime}=pi cos(pi x)`	`v^{prime}=1`

`g^{prime}(x)`	`=(vu^{prime}-uv^{prime})/v^2`
	`= ((x + 1) ⋅ pi cos(pi x)-sin (pi x))/(x + 1)^2`
`g^{prime}(1)`	`= (2 pi cos(pi)-sin(pi))/2^2`
	`= (2 pi(-1)-0)/4`
	`= -pi/2`

Calculus, MET1 2018 VCAA 1b

Let `f(x) = (e^x)/(cos(x))`.

Evaluate `f^{prime}(pi)`. (2 marks)

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`text(See Worked Solutions)`

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`f^{prime}(x) = (e^x)/(cos(x))`

`u`	`= e^x`	`v`	`= cos(x)`
`u^{prime}`	`= e^x`	`v^{prime}`	`=-sin(x)`

`f^{prime}(x)`	`= (u^{prime}v-uv^{prime})/(v^2)`
	`= (e^x · cos(x) + e^x sin(x))/(cos^2(x))`

`f^{prime}(pi)`	`= (e^pi · cospi + e^pi sinpi)/(cos^2 pi)`
	`= (e^pi(-1) + e^pi · 0)/((-1)^2)`
	`= -e^pi`

Calculus, MET1 2012 VCAA 1b

If `f(x) = x/(sin(x))`, find `f^{prime}(pi/2).` (2 marks)

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`1`

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`text(Using Quotient Rule:)`

`(h/g)^{prime}`	`= (h^{prime}g-h g^{prime})/g^2`
`f^{prime}(x)`	`= (1 xx sin (x)-x cos (x))/(sin x)^2`
`:. f^{prime}(pi/2)`	`= (sin (pi/2)-pi/2 xx cos (pi/2))/(sin(pi/2))^2`
	`= (1-0)/1^2`
	`= 1`

Calculus, MET1 2009 VCAA 1b

For `f(x) = (cos(x))/(2x + 2)` find `f prime (pi)`. (3 marks)

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`1/(2 (pi + 1)^2)`

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`text(Using Quotient Rule:)`

MARKER’S COMMENT: A majority of students did not substitute `x=pi` correctly.

`(g/h)^{prime}`	`= (g^{prime} h – gh^{prime})/h^2`
`f^{prime}(x)`	`= (-sin (x) (2x + 2)-2 cos (x))/(2x + 2)^2`
`:. f^{prime}(pi)`	`= (-sin (pi) (2pi + 2)-2 cos (pi))/(2pi + 2)^2`
	`= (0-2 (-1))/[2 (pi + 1)]^2`
	`= 2/(4(pi + 1)^2)`
	`= 1/(2 (pi + 1)^2)`

Calculus, MET1 2016 VCAA 1a

Let `y = (cos(x))/(x^2 + 2)`.

Find `(dy)/(dx)`. (2 marks)

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`(-x^2sin(x)-2sin(x)-2xcos(x))/((x^2 + 2)^2)`

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`text(Using Quotient Rule:)`

`(h/g)^{prime}`	`= (h^{prime}g-hg^{prime})/(g^2)`
`(dy)/(dx)`	`= (-sin(x)(x^2 + 2)-cos(x)(2x))/((x^2 + 2)^2)`
	`= (-x^2sin(x)-2sin(x)-2xcos(x))/((x^2 + 2)^2)`

Calculus, MET1 2007 VCAA 1

Let `f(x) = (x^3)/(sin(x))`. Find `f^{′}(x)`. (2 marks)

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`(3x^2sin(x)-x^3cos(x))/(sin^2(x))`

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`f(x) = (x^3)/(sin(x))`

`text(Using Quotient Rule:)`

`d/(dx)(u/v) = (vu^{prime}-uv^{prime})/(v^2)`

`:. f^{prime}(x) = (3x^2sin(x)-x^3cos(x))/(sin^2(x))`