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′(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′ = e^x`

`v = (x^2 – 3) \ \ => \ \ v′ = 2x`

`f′(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′(pi)`. (2 marks)

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`text(See Worked Solutions)`

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`f′(x) = (e^x)/(cos(x))`

`u`	`= e^x`	`v`	`= cos(x)`
`u′`	`= e^x`	`v′`	`= −sin(x)`

`f′(x)`	`= (u′v – uv′)/(v^2)`
	`= (e^x · cos(x) + e^x sin(x))/(cos^2(x))`

`f′(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′(x)`. (2 marks)
Evaluate `f′(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)′`	`= (h′g – hg′)/(g^2)`
`f′(x)`	`= ((1/x)x^2 – log_e(x)*2x)/(x^4)`
	`= (1 – 2log_e(x))/(x^3)`

ii.	`f′(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′`	`= 2`	`v′`	`= e^x`

`(dy)/(dx)`	`= (u′v – uv′)/(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)`