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Calculus, 2ADV C2 2023 MET1 1a

Let  \(y=\dfrac{x^2-x}{e^x}\).

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

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

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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, 2ADV C2 SM-Bank 7

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, 2ADV C2 2019 MET1 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, 2ADV C2 2018 HSC 11g

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

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

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

`text(Differentiate using quotient rule:)`

`u = e^x` `v = x + 1`
`u prime = e^x` `v prime = 1`
   
`(dy)/(dx)` `= (u prime v – u v prime)/v^2`
  `= (e^x(x + 1) – e^x ⋅ 1)/(x + 1)^2`
  `= (x e^x)/(x + 1)^2`

Calculus, 2ADV C2 2007 HSC 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`