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Calculus, 2ADV C1 2019 HSC 11c

Differentiate  `(2x + 1)/(x + 5)`.  (2 marks)

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`9/(x + 5)^2`

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`text(Using quotient rule:)`

`u=2x+1,`     `v=x+5`  
`u^{′} = 2,`     `v^{′} = 1`  
     
`y^{′}` `= (u^{′} v-v^{′} u)/v^2`
  `= (2(x + 5)-(2x + 1))/(x + 5)^2`
  `= (2x + 10-2x-1)/(x + 5)^2`
  `= 9/(x + 5)^2`

Calculus, 2ADV C1 2016 HSC 11b

Differentiate  `(x + 2)/(3x-4).`   (2 marks)

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`(-10)/(3x-4)^2`

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`y = (x + 2)/(3x-4)`

`text(Using the quotient rule:)`

`(g/h)^{′}` `= (g^{′} h-g h^{′})/h^2`
`y prime` `= (1 (3x-4)-(x + 2) · 3)/(3x-4)^2`
  `= (-10)/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c

Find  `f^{′}(x)`, where  `f(x) = (x^2 + 3)/(x-1).`   (2 marks)

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`((x-3) (x + 1))/(x-1)^2`

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`f(x) = (x^2 + 3)/(x-1)`

`text(Using the quotient rule:)`

`u` `= x^2 + 3` `\ \ \ \ \ \ v` `= x-1`
`u^{′}` `= 2x` `\ \ \ \ \ \ v^{′}` `= 1`
`f^{′}(x)` `= (u^{′} v-uv^{′})/v^2`
  `= (2x (x-1)-(x^2 + 3) xx 1)/(x-1)^2`
  `= (2x^2-2x-x^2-3)/(x-1)^2`
  `= (x^2-2x-3)/(x-1)^2`
  `= ((x-3) (x + 1))/(x-1)^2`

Calculus, 2ADV C1 2005 HSC 2bii

Differentiate with respect to `x`:

`x^2/(x-1).`   (2 marks)

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`(x(x-2))/(x-1)^2`

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`y = x^2/(x-1)`

`text(Using)\ \ dy/dx = (u^{′}v-uv^{′})/v^2:`

`u` `= x^2` `v` `= x-1`
`u^{′}` `= 2x` `v^{′}` `= 1`

 

`dy/dx` `= (2x(x-1)-x^2(1))/(x-1)^2`
  `= (2x^2-2x-x^2)/(x-1)^2`
  `= (x^2-2x)/(x-1)^2`
  `= (x(x-2))/(x-1)^2`

Calculus, 2ADV C1 2014 HSC 11c

Differentiate  `x^3/(x + 1)`.   (2 marks)

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`(x^2(2x + 3))/((x + 1)^2)`

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`y = (x^3)/(x + 1)`

`text(Using)\ \ \ dy/dx = (u^{′} v\ – uv^{′})/(v^2)`

`u` `=x^3` `\ \ \ \ \ v` `=(x+1)`
`u^{′}`  `=3x^2` `v^{′}` `=1`
`dy/dx` `= (3x^2 (x + 1)-x^3 (1))/((x + 1)^2)`
  `= (3x^3 + 3x^2-x^3)/((x + 1)^2)`
  `= (2x^3 + 3x^2)/((x + 1)^2)`
  `= (x^2 (2x + 3))/((x + 1)^2)`