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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`
Show Answers Only

`D`

Show Worked Solution
`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`

Filed Under: Differentiation and Integration, L&E Differentiation (Y12), Log Calculus, Log Calculus (Y12), Trig Differentiation (Y12) Tagged With: Band 3, smc-964-10-Differentiation, smc-964-40-Trig overlap, smc-967-20-Logs, smc-967-50-Chain Rule, smc-968-10-Sin, smc-968-60-Chain Rule

Calculus, 2ADV C4 2008 HSC 3b

  1. Differentiate  `log_e (cos x)`  with respect to  `x`.  (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.    (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `- tan x`
  2. `- log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`
Show Worked Solution
i.    `y` `= log_e (cos x)`
  `dy/dx` `= (- sin x)/(cos x)`
    `= – tan x`

 

ii.    `int_0^(pi/4) tan x\ dx`
  `= – [log_e (cos x)]_0^(pi/4)`
  `= – [log_e(cos (pi/4)) – log_e (cos 0)]`
  `= – [log_e (1/sqrt2) – log_e 1]`
  `= – [log_e (1/sqrt2) – 0]`
  `= – log_e (1/sqrt2)`
  `= 0.346…`
  `= 0.35\ \ text{(2 d.p.)}`

Filed Under: Differentiation and Integration, Log Calculus, Log Calculus (Y12), Trig Integration Tagged With: Band 3, Band 4, smc-1204-50-Diff then Integrate, smc-964-10-Differentiation, smc-964-40-Trig overlap, smc-964-50-Diff then integrate

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