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Calculus, 2ADV C1 EO-Bank 14 v1

Evaluate `f^{′}(1)`, where `f(x) = x^2 / sqrt(2x + 3)`. (4 marks)

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`9 / (5sqrt5)`

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

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

`= (2x)/(sqrt(2x + 3)) – (x^2)/(2x + 3)^(3/2)`

`= [2x(2x + 3) – x^2] / (2x + 3)^(3/2)`

`= (3x^2 + 6x) / (2x + 3)^(3/2)`

`f^{′}(1)` `= (3(1)^2 + 6(1)) / (2(1) + 3)^(3/2)`

`= 9 / (5sqrt5)`

Calculus, 2ADV C1 EO-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}`  to find  `f^{\prime}(x)`  if  `f(x)=x-3x^2`.   (2 marks)

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Show Worked Solution

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Calculus, 2ADV C1 2013 HSC 11b v1

Evaluate  `lim_(x->1) ((x-1)(x+2)^2)/(x^2+x-2)`.   (2 marks)

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 `3`

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`lim_(x ->1) ((x-1)(x+2)^2)/(x^2+x-2)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->1) ( (x -1)(x+2)^2)/( (x-1)(x+2)`

`=lim_(x->1) (x+2)`

`=3`

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 2023 HSC 7 MC v1

It is given that  \(y=f(g(x))\), where  \(f(2)=5\), \(f^{′}(2)=3\), \(g(4)=2\)  and  \(g^{′}(4)=-2\).

What is the value of \(y^{′}\) at  \(x=4\)?

  1. \(-6\)
  2. \(-2\)
  3. \(3\)
  4. \(6\)
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\(A\)

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\(y\) \(=f(g(x))\)  
\(y^{′}\) \(=f^{′}(g(4)) \times g^{′}(4)\)  
  \(=f^{′}(2) \times -2\)  
  \(=3 \times -2\)  
  \(=-6\)  

 
\(\Rightarrow A\)

Calculus, 2ADV C1 EO-Bank 1 MC v1

The derivative of  \((n^2-1) x^{3n-2}\)  can be expressed as

  1. \(3(n-1)(n^2-1) x^{3n-2}\)
  2. \(3(n-1)(n^2-1) x^{3(n-1)}\)
  3. \((3n-2) (n^2-1) x^{3(n-1)}\)
  4. \((3n-2) (n^2-1) x^{3n-2}\)
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\(C\)

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\(y\) \(=(n^2-1) x^{3n-2}\)  
\(y^{′}\) \(=(3n-2) (n^2-1) x^{3n-2-1)}\)  
  \(=(3n-2) (n^2-1) x^{3(n-1)}\)  

 
\(\Rightarrow C\)

Calculus, 2ADV C1 EO-Bank 7

Differentiate  `2x(1-4x)^5`  with respect to `x`.   (2 marks)

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`y^{′}=2(1-4x)^4(1-24x)`

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`y=2x(1-4x)^5`

`text{Using the product and chain rules:}`

`y^{′}` `=2 xx (1-4x)^5-40x(1-4x)^4`  
  `=2(1-4x)^4(1-4x-20x)`  
  `=2(1-4x)^4(1-24x)`  

Calculus, 2ADV C1 EO-Bank 3

  1.  Use differentiation by first principles to find \(y^{′}\), given  \(y = 4x^2 - 5x + 4\).   (2 marks)

  2.  Find the equation of the tangent to the curve when  \(x = 3\).   (1 mark)

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  1.  `y^{′} = 8x-5`
  2.  `y = 19x-32`
Show Worked Solution
i.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((4(x + h)^2-5(x + h) + 4)-(4x^2-5x + 4))/h`
    `= lim_(h->0)(4x^2 + 8xh + 4h^2-5x-5h + 4-4x^2+5x-4)/h`
    `= lim_(h->0)(8xh + 4h^2-5h)/h`
    `= lim_(h->0)(h(8x-5 + 4h))/h`

 
`:.\ y^{′} = 8x-5`
 

ii.   `text(When)\ \ x = 3, y = 25`

`y^{′} = 24-5 = 19`
 

`:. y-25` `= 19(x-3)`
`y` `= 19x-32`