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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 EQ-Bank 3

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

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

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  1.  `y^{′} = 4x+5`
  2.  `y = 9x-2`
Show Worked Solution
i.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((2(x + h)^2 +5(x + h))-(2x^2+5x))/h`
    `= lim_(h->0)(2x^2 + 4xh + 2h^2+5x+5h-2x^2-5x)/h`
    `= lim_(h->0)(4xh + 2h^2+5h)/h`
    `= lim_(h->0)(h(4x+5 +2h))/h`

 
`:.\ y^{′} = 4x+5`
 

ii.   `text(When)\ \ x = 1, y = 7`

`y^{′} = 4+5 = 9`
 

`:. y-7` `= 9(x-1)`
`y` `= 9x-2`

Calculus, 2ADV C1 EQ-Bank 4

Use differentiation by first principles to find \(y^{\prime}\) given  \(y=\dfrac{5}{x}\).   (3 marks)

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\(\text{See worked solutions}\)

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\(\dfrac{d}{d x}\left(\dfrac{5}{x}\right)\) \(=\displaystyle \lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\)
  \(=\displaystyle \lim _{h \rightarrow 0}\dfrac{\dfrac{5}{x+h}-\dfrac{5}{x}}{h}\)
  \(=\displaystyle \lim _{h \rightarrow 0} \dfrac{\dfrac{5 x-5(x+h)}{x(x+h)}}{h}\)
  \(=\displaystyle \lim _{h \rightarrow 0} \dfrac{\dfrac{-5 h}{x(x+h)}}{h}\)
  \(=\displaystyle \lim _{h \rightarrow 0} \dfrac{-5}{x(x+h)}\)
  \(=\dfrac{-5}{x^2}\)

Calculus, 2ADV C1 EQ-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)=5x^2-2x`.   (2 marks)

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`f(x)=5x^2-2x`

`f^{′}(x)` `= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`  
  `= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`  
  `= \lim_{h->0} \frac{10xh+h^2-2h}{h}`  
  `= \lim_{h->0} \frac{h(10x+h-2)}{h}`  
  `= \lim_{h->0} 10x+h-2`  
  `=10x-2`  

Show Worked Solution

`f(x)=5x^2-2x`

`f^{′}(x)` `= \lim_{h->0} \frac{5(x+h)^2-2(x+h)-(5x^2-2x)}{h}`  
  `= \lim_{h->0} \frac{5x^2+10xh+h^2-2x-2h-5x^2+2x}{h}`  
  `= \lim_{h->0} \frac{10xh+h^2-2h}{h}`  
  `= \lim_{h->0} \frac{h(10x+h-2)}{h}`  
  `= \lim_{h->0} 10x+h-2`  
  `=10x-2`  

Calculus, 2ADV C1 EQ-Bank 2

When differentiating  `f(x) = 3-2x-x^2`  from first principles, a student began the solution as shown below.

Complete the solution.  (2 marks)

   `f^{′}(x) = lim_(h->0) (f(x + h)-f(x))/h`

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`f^{′}(x) = -2x-2`

Show Worked Solution
   `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) (3-2(x + h) – (x+h)^2-(3-2x-x^2))/h`
    `= lim_(h->0) (3-2x-2h-x^2-2hx-h^2 – 3 + 2x + x^2)/h`
    `= lim_(h->0) (-2h-2hx-h^2)/h`
    `= lim_(h->0) (h(-2x-2-h))/h`

 
`:.\ f^{′}(x) = -2x-2`

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`