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Calculus, 2ADV C2 EQ-Bank 26

Show that  \(\dfrac{d}{d x}(\cot x)=-\operatorname{cosec}^2 x\).   (2 marks)

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\(\text{See Worked Solutions}\)

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\(\cot\,x=\dfrac{\cos\,x}{\sin\,x}\)

\(\text{Using the quotient rule:}\)

\(\dfrac{d}{d x}\left(\dfrac{\cos x}{\sin x}\right)\) \(=\dfrac{-\sin\, x\,\sin\, x-\cos\, x\, \cos \,x}{\sin ^2 x}\)  
  \(=\dfrac{-\sin ^2 x-\cos ^2 x}{\sin ^2 x}\)  
  \(=-\left(\dfrac{1}{\sin ^2 x}\right)\)  
  \(=-\operatorname{cosec}^2 x\)  

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) = (vu^{\prime}-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 2017 HSC 11c

Differentiate  `(sin x)/x`.   (2 marks)

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`(x cos x-sin x)/x^2`

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`y = (sin x)/x`

`text(Let)\ \ u` `=sin x` `u^{prime}` `= cos x`
`v` `=x` `v ^{prime}` `=1`

 

`(dy)/(dx)` `= (u^{prime}v-u v^{prime})/v^2`
  `= (x cos x-sin x)/x^2`

Calculus, 2ADV C2 2008 HSC 2aiii

Differentiate with respect to  `x`:

`sinx/(x+4)`.   (2 marks) 

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`(cosx (x+4)-sin x)/((x + 4)^2)`

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`y = sinx/(x + 4)`

`u` `= sinx` `\ \ \ \ \ u^{prime}` `= cos x`
`v` `= x + 4` `v^{prime}` `= 1`

 

`dy/dx` `= (u^{prime}v-uv^{prime})/v^2`
  `= (cos x (x + 4)-sin x)/(x+4)^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` `v = x`
  `\ \ u^{prime} =-sin 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\ \ \ \ \ \ \ ` `v = cosx`
  `u ^ {prime} = 1\ \ \ \ \ \ \ ` `v^{prime} =-sin x`

 

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

 
`=>  A`