smc-964-10-Differentiation

Calculus, 2ADV C2 EQ-Bank 4 MC

If `f(x)=log_2(x^(2x))`, which expression is equal to `f^(′)(x)`?

`2/(x^(2x)ln2`
`2/ln2 + 2log_2x`
`log_2x+2/ln2`
`2/ln2 xx log_2(x^(2x-1))`

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`B`

Show Worked Solution

`f(x)`	`=log_2(x^(2x))`
	`=2x log_2x`
	`=(2x lnx)/ln2`

`f^(′)(x)`	`=1/ln2 (2x*1/x + 2lnx)`
	`=2/ln2 + (2lnx)/ln2`
	`=2/ln2 + 2log_2x`

`=> B`

Calculus, 2ADV C2 SM-Bank 8

Let `y= (x + 5) log_e (x)`.

Find `(dy)/(dx)` when `x = 5`. (2 marks)

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`log_e 5 +2`

Show Worked Solution

`(dy)/(dx)`	`= 1 xx log_e x + (x + 5) * (1)/(x)`
	`= log_e x + (x + 5)/(x)`

`:. dy/dx|_(x=5)=log_e 5 +2`

Calculus, 2ADV C2 SM-Bank 6

Differentiate with respect to `x`:

`log_e x^x`. (2 marks)

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`1 + log_ex`

Show Worked Solution

`y`	`=log_e x^x`
	`=xlog_ex`
`dy/dx`	`=x*1/x + log_ex`
	`=1 + log_ex`

Calculus, 2ADV C2 EQ-Bank 1

Differentiate `log_2 x^2` with respect to `x`. (2 marks)

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`2/(xln2)`

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TIP: The new Advanced reference sheet can be used here!

`y`	`= log_2 x^2`
`(dy)/(dx)`	`= {:d/(dx):} ((lnx^2)/(ln2))`
	`= 1/(ln2) · d/(dx)(ln x^2)`
	`= 1/(ln2) · (2x)/(x^2)`
	`= 2/(xln2)`

Calculus, 2ADV C4 2019 HSC 13c

Differentiate `(ln x)^2`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find `int(ln x)/x\ dx`. (1 mark)

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

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`(2 ln x)/x`
`1/2 (ln x)^2 + C`

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i.	`y`	`= (ln x)^2`
	`(dy)/(dx)`	`= 2 ⋅ 1/x ⋅ ln x`
		`= (2 ln x)/x`

♦ Mean mark part (ii) 49%.

ii.	`int (ln x)/x\ dx`	`=1/2 int (2 ln x)/x dx`
		`= 1/2 (ln x)^2 +C`

Calculus, 2ADV C4 2018 HSC 5 MC

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

`cos (1/x)`
`cos (ln x)`
`cos ((ln x)/x)`
`(cos (ln x))/x`

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`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`

Calculus, 2ADV C2 2017 HSC 11d

Differentiate `x^3 ln x`. (2 marks)

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`x^2 (3 ln\ x + 1)`

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`y = x^3 ln\ x`

`text(Using the product rule:)`

`(dy)/(dx)`	`= 3x^2 * ln\ x + x^3 * 1/x`
	`= x^2 (3 ln\ x + 1)`

Calculus, 2ADV C2 2015 HSC 11f

Differentiate `y = (x + 4) ln\ x`. (2 marks)

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`ln\x + 4/x +1`

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`y = (x + 4) ln\ x`

`text(Using the product rule)`

`(dy)/(dx)`	`= d/(dx) (x + 4) * ln x + (x + 4) d/(dx) ln\ x`
	`= ln x + (x + 4) 1/x`
	`= ln x + 4/x + 1`

Calculus, 2ADV C2 2008 HSC 2aii

Differentiate with respect to `x`:

`x^2 log_e x` (2 marks)

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`x + 2x log_e x`

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`y`	`= x^2 log_e x`
`dy/dx`	`= x^2 * 1/x + 2x * log_e x`
	`= x + 2x log_e x`

Calculus, 2ADV C4 2008 HSC 3b

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

--- 4 WORK AREA LINES (style=lined) ---
Hence, or otherwise, evaluate `int_0^(pi/4) tan x\ dx`. (2 marks)

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

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`- tan x`
`- 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.)}`

Calculus, 2ADV C2 2011 HSC 1d

Differentiate `ln(5x+2)` with respect to `x`. (2 marks)

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`5/(5x+2)`

Show Worked Solutions

`y`	`=ln(5x+2)`
`dy/dx`	`=5/(5x+2)`

Calculus, 2ADV C2 2012 HSC 12ai

Differentiate with respect to `x`

`(x-1)log_ex` (2 marks)

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`log_ex+1-1/x`

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`y`	`=(x-1)log_ex`
`dy/dx`	`=1(log_ex)+(x-1)1/x`
	`=log_ex+1-1/x`