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Calculus, 2ADV C2 2006 HSC 2aii

Differentiate  `sin x/(x + 1)`  with respect to `x`.  (2 marks)

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`dy/dx = {cos x (x + 1)-sin x} / (x + 1)^2`

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`y = sin x / (x + 1)`

`text(Using)\ \ d/dx (u/v) = (u^{\prime} v-uv^{\prime})/v^2`

`u` `= sin x` `v` `= x + 1`
`u^{\prime}` `= cos x` `\ \ \ v^{\prime}` `= 1`

 

`:.dy/dx = {cos x (x + 1)-sin x} / (x + 1)^2`

Calculus, 2ADV C2 2010 HSC 2a

Differentiate  `cosx/x`  with respect to  `x`.   (2 marks) 

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 `(-x sinx – cos x)/(x^2)`

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MARKER’S COMMENT: A significant number of students did not know the quotient rule and many found assistance by writing out `u,\ u prime,\ v,\ v prime\ ` before going into calculations.

`y = cos x/x`

`text(Let)` `\ \ u = cos x` `u prime = – sin x`
  `\ \ v = x` `v prime = 1`

 

`text(Using quotient rule:)`

`dy/dx` `= (u prime v – u v prime)/(v^2)`
  `= (-sinx *x  – cos x*1)/x^2`
  `= (-x sin x – cos x)/(x^2)`

Calculus, 2ADV C2 2011 HSC 4a

Differentiate  `x/sinx`  with respect to  `x`.   (2 marks)

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 `(sin x\ – x cos x)/(sin^2x)`

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`y = x/sinx`

`u = x` `\ \ \ \ \ u prime = 1`
`v = sin x` `\ \ \ \ \ v prime = cos x`

 

`text(Using)\ \ d/dx (uv) = (u prime v\ – uv prime)/(v^2),`

`dy/dx` `= (1 * sinx \ – x * cos x)/((sin x)^2)`
  `= (sin x\ – x cos x)/(sin^2x)`

Calculus, 2ADV C2 2013 HSC 4 MC

What is the derivative of  `x/cosx`?

  1. `(cosx+xsinx)/(cos^2 x)`  
  2. `(cosx-xsinx)/(cos^2 x)`  
  3. `(xsinx-cosx)/(cos^2 x)`
  4. `(-xsinx-cosx)/(cos^2 x)`
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`A`

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`y = x/cosx`

`text(Let)\ \ \ \ ` `u = x\ \ \ \ \ \ \ ` `u prime = 1`
  `v = cosx\ \ \ \ \ \ \ ` `v prime =-sin x`

 

`:.\ dy/dx` `= (vu prime\-uv prime)/v^2`
  `= (cosx  1-x (- sinx))/(cosx)^2`
  `= (cosx + xsinx)/(cos^2x)`

 
`=>  A`