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Calculus, 2ADV C4 2022 HSC 18

  1. Differentiate  `y=(x^(2)+1)^(4)`.  (2 marks)

  2. Hence, or otherwise, find `int x(x^(2)+1)^(3)dx`.  (1 mark)

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  1.  `dy/dx=8x(x^2+1)^3`
  2.  `1/8(x^2+1)^4+C`
Show Worked Solution

a.   `y=(x^2+1)^4`

`text{Using chain rule:}`

`dy/dx` `=4 xx 2x(x^2+1)^3`  
  `=8x(x^2+1)^3`  

 

b.    `intx(x^2+1)^3\ dx` `=1/8 int8x(x^2+1)^3\ dx`
    `=1/8(x^2+1)^4+C`

Calculus, 2ADV C4 2011 HSC 4d

  1. Differentiate  `y=sqrt(9 - x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (6x)/sqrt(9 - x^2)\ dx`.    (2 marks)

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  1. `- x/sqrt(9\ – x^2)`
  2. `-6 sqrt(9\ – x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(9-x^2)^(1/2)`.
i.    `y` `= sqrt(9 – x^2)`
    `= (9 – x^2)^(1/2)`

 

`dy/dx` `=1/2 xx (9 – x^2)^(-1/2) xx d/dx (9 – x^2)`
  `= 1/2 xx (9 – x^2)^(-1/2) xx -2x`
  `= – x/sqrt(9 – x^2)`

 

ii.    `int (6x)/sqrt(9 – x^2)\ dx` `= -6 int (-x)/sqrt(9 – x^2)\ dx`
    `= -6 (sqrt(9 – x^2)) + C`
    `= -6 sqrt(9 – x^2) + C`