smc-739-50-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 2023 VCAA 1a

Let \(y=\dfrac{x^2-x}{e^x}\).

Find and simplify \(\dfrac{dy}{dx}\). (2 marks)

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\(-\Bigg(\dfrac{x^2-3x+1}{e^x}\Bigg) \)

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\(\text{Using the quotient rule:}\)

\(\dfrac{dy}{dx}\)	\(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\)
	\(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\)
	\(=\dfrac{-x^2+3x-1}{e^x}\)

Calculus, MET1-NHT 2018 VCAA 1a

Let `f(x) = (e^x)/((x^2-3))`.

Find `f^{prime}(x)`. (2 marks)

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`{e^x(x^2-2x-3)}/{(x^2-3)^2}`

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`text(Let) \ \ u = e^x \ \ => \ \ u^{prime} = e^x`

`v = (x^2-3) \ \ => \ \ v^{prime} = 2x`

`f^{prime}(x)`	`= {e^x(x^2-3)-2x e^x}/{(x^2-3)^2}`
	`= {e^x(x^2-2x-3)}/{(x^2-3)^2}`

Calculus, MET1-NHT 2019 VCAA 1a

Let `y = (2e^(2x)-1)/e^x`.

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

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`(dy)/(dx) = 2e^x + e^(-x)`

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`text(Method 1)`

`y`	`= 2e^x-e^(-x)`
`(dy)/(dx)`	`= 2e^x + e^(-x)`

`text(Method 2)`

`(dy)/(dx)`	`= (4e^(2x) ⋅ e^x-(2e^(2x)-1) e^x)/(e^x)^2`
	`= (4e^(3x)-2e^(3x) + e^x)/e^(2x) `
	`= (2e^(2x) + 1)/e^x`

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 2015 VCAA 1b

Let `f(x) = (log_e(x))/(x^2)`.

Find `f^{prime}(x)`. (2 marks)

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Evaluate `f^{prime}(1)`. (1 mark)

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`(1 – 2log_e(x))/(x^3)`
`1`

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i. `text(Using Quotient Rule:)`

`(h/g)^{prime}`	`= (h^{prime}g-hg^{prime})/(g^2)`
`f^{prime}(x)`	`= ((1/x)x^2-log_e(x)*2x)/(x^4)`
	`= (1-2log_e(x))/(x^3)`

ii.	`f^{prime}(1)`	`= (1-2log_e(1))/(1^3)`
		`= 1`

Calculus, MET1 2007 ADV 2ai

Differentiate with respect to `x`:

`(2x)/(e^x + 1).` (2 marks)

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`(2(e^x + 1-xe^x))/((e^x + 1)^2)`

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`y = (2x)/(e^x + 1)`

`u`	`= 2x`	`v`	`= e^x + 1`
`u^{\prime}`	`= 2`	`v^{\prime}`	`= e^x`

`(dy)/(dx)`	`= (u^{\prime}v – uv^{\prime})/(v^2)`
	`= (2(e^x + 1)-2x(e^x))/((e^x + 1)^2)`
	`= (2e^x + 2-2x · e^x)/((e^x + 1)^2)`
	`= (2(e^x + 1-xe^x))/((e^x + 1)^2)`