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Calculus, 2ADV C1 2019 HSC 11c v1

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

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`-25/(3x-4)^2`

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`text(Using quotient rule:)`

`u=4x+3,`     `v=3x-4`  
`u^{′} = 4,`     `v^{′} = 3`  
     
`y^{′}` `= (u^{′} v-v^{′} u)/v^2`
  `= (4(3x-4)-3(4x+3))/(3x-4)^2`
  `= (12x-16-12x-9)/(3x-4)^2`
  `= -25/(3x-4)^2`

Calculus, 2ADV C1 2015 HSC 12c v1

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

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

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

`text(Using the quotient rule:)`

`u` `= 2x^2-3x` `\ \ \ \ \ \ v` `= 2-x`
`u^{′}` `= 4x-3` `\ \ \ \ \ \ v^{′}` `= -1`
`f^{′}(x)` `= (u^{′} v-uv^{′})/v^2`
  `= ((4x-3)(2-x)-(2x^2-3x) xx -1)/(2-x)^2`
  `= (-2x^2 + 8x-6)/(x-2)^2`
  `= (-2(x^2-4x+3)/(x-2)^2`
  `= (-2(x-3) (x-1))/(x-2)^2`

Calculus, 2ADV C1 EO-Bank 11 MC v1

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-1)=7,\ g(-1)=5\)  and  \(f^{′}(-1)=-4,\ g^{′}(-1)=-2\).

The gradient of the graph  \(y=\dfrac{f(x)}{g(x)}\)  at the point where  \(x=-1\)  is

  1. \(-\dfrac{6}{49}\)
  2. \(\dfrac{6}{49}\)
  3. \(\dfrac{6}{25}\)
  4. \(-\dfrac{6}{25}\)
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\(D\)

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\(\text{Using the Quotient Rule when}\ \ x=-1:\)

\(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\) \(=\dfrac{g(x)f^{′}(x)-f(x)g^{′}(x)}{g(x)^2}\)
  \(=\dfrac{g(-1)f^{′}(-1)-f(-1)g^{′}(-1)}{g(-1)^2}\)
  \(=\dfrac{5 \times -4-7 \times -2}{5^2}\)
  \(=-\dfrac{6}{25}\)

 
\(\Rightarrow D\)

Calculus, 2ADV C1 2022 HSC 5 MC

Let  `h(x)=(f(x))/(g(x))`, where

`{:[f(1)=2, qquad f^{′}(1)=4],[g(1)=8, qquad g^{′}(1)=12]:}`

What is the gradient of the tangent to the graph of  `y=h(x)`  at  `x=1` ?

  1. `-8`
  2. `\ \ \ 8`
  3. `- 1/8`
  4. `\ \ \ 1/8`
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`D`

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`text{Using the quotient rule:}`

`h^{′}(1)` `=(g(1)\ f^{′}(1)-f(1)\ g^{′}(1))/[g(1)]^2`  
  `=(8xx4-2xx12)/(8^2)`  
  `=(32-24)/64`  
  `=1/8`  

 
`=>D`

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) prime` `= (g prime h – g h prime)/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 prime v\ – uv prime)/(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)`