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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 EQ-Bank 19

  1. Prove that  \(\cos x+\sin x\, \tan x=\sec x\).   (1 mark)

  2. Hence, or otherwise, find  \(\dfrac{d}{dx}\left(\dfrac{1}{\cos x+\sin x \,\tan x}\right)\).   (2 marks)

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a.    \(\text{Proof (See worked solutions)}\)

b.    \(-\sin x\)

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a.    \(\text{Prove}\ \ \cos x+\sin x\, \tan x=\sec x\)

\(\text{LHS}\) \(=\cos x+\sin x \cdot \dfrac{\sin x}{\cos x}\)
  \(=\dfrac{\cos ^2 x+\sin ^2 x}{\cos x}\)
  \(=\dfrac{1}{\cos x}\)
  \(=\sec x=\text{RHS}\)

 

b.     \(\begin{array}{r} \dfrac{d}{d x}\end{array} \left(\dfrac{1}{\cos x+\sin x\, \tan x}\right)\) \(=\dfrac{d}{d x}\left(\dfrac{1}{\sec x}\right)\)
    \(=\begin{array}{r}\dfrac{d}{dx}\end{array} \left(\cos x\right)\)
    \(=-\sin x\)

Calculus, 2ADV C2 2022 HSC 25

Let  `f(x)=sin(2x)`.

Find the value of  `x`, for  `0 < x < pi`, for which  `f^(')(x)=-sqrt3`  AND  `f^('')(x)=2`.   (3 marks)

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`(7pi)/12`

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`f^(′)(x)=2cos(2x)`

`2cos(2x)` `=-sqrt3`
`cos(2x)` `=-sqrt3/2`
`2x` `=pi-pi/6,\ \ pi+pi/6`
  `=(5pi)/6,\ \ (7pi)/6`
`x` `=(5pi)/12,\ \ (7pi)/12`

 
`f^(″)(x)=-4sin(2x)`

`-4sin(2x)` `=2`
`sin(2x)` `=-1/2`
`2x` `=pi+pi/6,\ \ 2pi-pi/6`
  `=(7pi)/6,\ \ (11pi)/6`
`x` `=(7pi)/12,\ \ (22pi)/12`

 
`:.x=(7pi)/12\ \ text{(satisfies both equations)}`


Mean mark 53%.

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 EQ-Bank 28

The function  `f(theta) = sin^3(2 theta)`.

If  `f^{prime}(theta) = 6 cos(2 theta)-6 cos^n (2 theta)`, find the value of `n`.   (2 marks)

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`3`

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`f(theta)= sin^3(2 theta)= (sin(2theta))^3`

`f^{prime}(theta)` `= 3 xx 2cos(2 theta) xx sin^2(2 theta)`
  `= 6 cos(2 theta)(1-cos^2(2 theta))`
  `= 6 cos (2 theta)-6 cos^3(2 theta)`

 
`:. \ n = 3`

Calculus, 2ADV C2 EQ-Bank 14

Differentiate with respect to `x`: 

`e^(tan(2x))`   (2 marks)

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 `2 sec^2(2x)* e^(tan(2x))`

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`y` `=e^(tan(2x))`
`dy/dx` `= d/(dx)tan(2x) xx e^(tan(2x))`
  `= 2 sec^2(2x)* e^(tan(2x))`

Calculus, 2ADV C2 2019 MET1 1b

Let  `f(x) = x^2 cos(3x)`.
 
Find  `f^{prime}(pi/3)`.   (2 marks)

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`-(2pi)/3`

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`f(x)` `= x^2 cos 3x`
`f^{prime}(x)` `= x^2 ⋅ 3(-sin 3x) + 2x cos 3x`
`f^{prime}(pi/3)`  `= (pi/3)^2 ⋅ 3 (-sin pi) + 2 (pi/3) cos pi`
  `= -(2pi)/3`

Calculus, 2ADV C2 2018 HSC 5 MC

What is the derivative of  `sin(ln x),` where  `x > 0`?

  1. `cos (1/x)`
  2. `cos (ln x)`
  3. `cos ((ln x)/x)`
  4. `(cos (ln x))/x`
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`D`

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`y` `= sin (ln x)`
`(dy)/(dx)` `= cos (ln x) xx d/(dx) (ln x)`
  `= cos (ln x) xx 1/x`
  `= (cos (ln x))/x`

  
 `=>  D`

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 2016 HSC 5 MC

What is the derivative of  `ln (cos x)?`

  1. `-sec x`
  2. `-tan x`
  3. `sec x`
  4. `tan x`
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`B`

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`y` `= ln (cos x)`
`(dy)/(dx)` `= (-sin x)/(cos x)`
  `= -tan x`

  
`=>  B`

Calculus, 2ADV C2 2007 HSC 2aii

Differentiate with respect to `x`:

`(1 + tan x)^10`.   (2 marks)

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`10 sec^2 x \ (1 + tan x)^9`

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`y = (1 + tan x)^10`

`(dy)/(dx)` `= 10 (1 + tan x)^9 xx d/(dx) (1+tan x)`
  `= 10 sec^2 x \ (1 + tan x)^9`

Calculus, 2ADV C2 2004 HSC 3aii

Differentiate with respect to  `x`:

`(1 + sin x)^5`.   (2 marks)

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`5 cos x\ (1 + sinx)^4`

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`y` `= (1 + sinx)^5`
`dy/dx` `= 5 (1 + sinx)^4 xx d/(dx)(sinx)`
  `= 5 cos x (1 + sinx)^4`

Calculus, 2ADV C2 2006 HSC 2ai

Differentiate  `x tan x`  with respect to `x`.  (2 marks)

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`dy/dx = x  sec^2 x + tan x `

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`y = x tan x`

`text(Using product rule)`

`d/dx (uv)` `=uv ^{prime}+ u^{prime}v`
`:.dy/dx` `=x xx sec^2 x+ 1 xxtan x `
  `= x sec^2 x + tan x`

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 2010 HSC 1e

Differentiate  `x^2 tan x`  with respect to  `x`.   (2 marks)

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`2x tanx + x^2 sec^2 x`

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COMMENT: There is no necessity to take out the common factor `x` in this case.

`y = x^2 tan x`

`text(Using product rule:)`

`d/dx (uv)` ` = u^{prime} v + u v^{prime}`
`dy/dx` `=2x tanx + x^2 sec^2 x`

Calculus, 2ADV C2 2012 HSC 12aii

Differentiate with respect to  `x`. 

`(cos x)/(x^2)`.   (2 marks)

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`(-x sin x-2 cos x)/(x^3)`

 

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`y = cosx/(x^2)`

`u = cos x` `\ \ \ \ \ \ v = x^2`
`u^{prime} =-sin x` `\ \ \ \ \ \ v^{prime}= 2x`

 

`text(Using the quotient rule,)` 

`dy/dx` `= (vu^{prime}\-uv^{prime})/(v^2)`
  `= (x^2*-sin x-cos x*2x )/(x^4)`
  `= (-x sin x-2 cos x)/(x^3)`

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`