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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 EQ-Bank 2

Differentiate with respect to  `x`: 

`e^(tan(2x))`   (2 marks)

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 `2 sec^2(2x)* e^(tan(2x))`

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`y` `=e^(tan(2x))`
`dy/dx` `= d/(dx)tan(2x) xx e^(tan(2x))`
  `= 2 sec^2(2x)* e^(tan(2x))`

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 C4 2016 HSC 12d

  1. Differentiate  `y = xe^(3x)` . (1 mark)

  2. Hence find the exact value of  `int_0^2 e^(3x) (3 + 9x)\ dx`.  (2 marks)

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  1. `e^(3x) (1 + 3x)`
  2. `6e^6`
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i.  `y = xe^(3x)`

`text(Using product rule:)`

`(dy)/(dx)` `= x · 3e^(3x) + 1 · e^(3x)`
  `= e^(3x) (1 + 3x)`

 

ii.  `int_0^2 e^(3x) (3 + 9x)\ dx`

`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6 – 0)`

`= 6e^6`

Calculus, 2ADV C2 2015 HSC 11e

Differentiate  `(e^x + x)^5`.  (2 marks)

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`5 (e^x + 1) (e^x + x)^4`

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`y` `= (e^x + x)^5`
 `(dy)/(dx)` `= 5 (e^x + x)^4 xx d/(dx) (e^x + x)`
  `= 5 (e^x + x)^4 xx (e^x + 1)`
  `= 5 (e^x + 1) (e^x + x)^4`

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`

Calculus, 2ADV C2 2012 HSC 11d

Differentiate    `(3+e^(2x))^5`.    (2 marks) 

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`10e^(2x)(3+e^(2x))^4`

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`y=(3+e^(2x))^5`

`(dy)/dx` `=5(3+e^(2x))^4 xx  d/(dx)(3+e^(2x))`
  `=5(3+e^(2x))^4 xx 2e^(2x)`
  `=10e^(2x)(3+e^(2x))^4`

 

Calculus, 2ADV C2 2013 HSC 11d

Differentiate  `x^2e^x`    (2 marks)

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 `xe^x(x+2)`

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`text{Using the product rule}`

`text(Let)\ \ u=x^2,` `\ \ \ \ \ \ u^{\prime}=2x`
`text(Let)\ \ v=e^x,` `\ \ \ \ \ \ v^{\prime}=e^x`
`{d(uv)}/dx` `=u prime v+v prime u`
  `=2x e^x +x^2 e^x `
  `=xe^x(x+2)`