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Calculus, 2ADV C2 EQ-Bank 30

Differentiate with respect to \(x\) :

\(f(x)=\log _e\left(\dfrac{x^3}{3-2 x}\right)\)   (3 marks)

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\(f^{\prime}(x)=\dfrac{9-4 x}{x(3-2 x)}\)

Show Worked Solution

\begin{align}
\begin{aligned}
\text {Let} \ \ & u=x^3 & u^{\prime}=3 x^2 \\
& v=3-2 x & v^{\prime}=-2
\end{aligned}
\end{align}

\(f^{\prime}(x)\) \(=\dfrac{\dfrac{v u^{\prime}-u v^{\prime}}{v^2}}{\frac{x^3}{3-2 x}}\)
  \(=\dfrac{(3-2 x) 3 x^2-x^3(-2)}{(3-2 x)^2} \times \dfrac{3-2 x}{x^3}\)
  \(=\dfrac{x^2(9-6 x+2 x)}{(3-2 x)} \times \dfrac{1}{x^3}\)
  \(=\dfrac{9-4 x}{x(3-2 x)}\)

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-30-\(\log_e x\), smc-7128-40-Quotient Rule, smc-967-20-Logs, smc-967-40-Quotient Rule

Calculus, 2ADV C2 EQ-Bank 7 MC

Given the function  \(f(x)=\log _{10} x^x\), which of the following expressions is equal to \(f^{\prime}(x)\) ?

  1. \(\log _e 10+\log _e x\)
  2. \(\dfrac{\log _e 10+1}{\log _e 10}\)
  3. \(\dfrac{1}{\log _e 10}+\log _x 10\)
  4. \(\dfrac{1}{\log _e x}+\log _{10} x\)
Show Answers Only

\(B\)

Show Worked Solution

\(f(x)=\log _{10} x^x=x \log _{10} x\)

\(\text{Using product rule:}\)

\(f^{\prime}(x)\) \(=x \cdot \dfrac{1}{x \cdot \ln 10}+1 \cdot \log _{10} x\)
  \(=\dfrac{1}{\ln 10}+\log _{10} x\)
  \(=\dfrac{1}{\ln 10}+\dfrac{\ln x}{\ln 10}\)
  \(=\dfrac{\ln x+1}{\ln 10}\)

 
\(\Rightarrow B\)

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 5, smc-7128-20-\(\large a^x\), smc-7128-35-Product Rule, smc-967-15-Exponentials (base a), smc-967-30-Product Rule

Calculus, 2ADV C3 2024 MET1 1b

Let  \(f(x)=\log _e\left(x^3-3 x+2\right)\).

Find  \(f^{\prime}(3)\)   (2 marks)

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\(\dfrac{6}{5}\)

Show Worked Solution

\(f(x)\) \(=\log_{e}(x^3-3x+2)\)
\(f^{\prime}(x)\) \(=\dfrac{3x^2-3}{x^3-3x+2}\)
\(f^{\prime}(3)\) \(=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}=\dfrac{6}{5}\)

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-30-\(\log_e x\), smc-967-20-Logs

Calculus, 2ADV C3 2024 MET1 1a

Let  \(y=e^x \cos\,3 x\).

Find  \(\dfrac{d y}{d x}\)   (2 marks)

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\(e^x (\cos(3x)-3\sin(3x))\)

Show Worked Solution

\(y\) \(=e^x \cos(3x)\)
\(\dfrac{dy}{dx}\) \(=e^x.(-3\sin(3x))+\cos(3x).e^x\)
  \(=e^x(\cos(3x)-3\sin(3x))\)

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-35-Product Rule, smc-7128-70-Trig Overlap, smc-7129-20-Cos, smc-7129-40-Product Rule, smc-7129-70-Log/Exp Overlap, smc-967-10-Exponentials (base e), smc-967-30-Product Rule, smc-967-80-Trig Overlap

Calculus, 2ADV C2 2023 MET1 1a

Let  \(y=\dfrac{x^2-x}{e^x}\).

Find and simplify \(\dfrac{dy}{dx}\).   (2 marks)

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\(\dfrac{-x^2+3x-1}{e^x}\)

Show Worked Solution

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

\(\dfrac{dy}{dx}\) \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\)
  \(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\)
  \(=\dfrac{-x^2+3x-1}{e^x}\)

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-10-\(\large e^x\), smc-7128-40-Quotient Rule, smc-965-10-Differentiation (base e), smc-965-30-Indefinite integrals, smc-967-10-Exponentials (base e), smc-967-40-Quotient Rule

Calculus, 2ADV C2 EQ-Bank 27

Differentiate  `pi^(2x)`.   (2 marks)

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`2log_e(pi) * pi^(2x)`

Show Worked Solution
COMMENT: `pi` is a constant. i.e. differentiate `a^(f(x))`.
`y` `=pi^(2x)`
`dy/dx` `=log_e(pi) * 2 * pi^(2x)`
  `=2log_e(pi) *pi^(2x)`

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-20-\(\large a^x\), smc-7128-50-Chain Rule, smc-965-20-Differentiation (base a), smc-967-15-Exponentials (base a), smc-967-50-Chain Rule, smc-967-60-New Reference Sheet

Calculus, 2ADV C2 EQ-Bank 5 MC

If  `f(x)=e^(g(x^(2)))`, where `g` is a differentiable function, then `f^(′)(x)` is equal to

  1. `2xe^(g(x^(2)))`
  2. `2xg(x^(2))e^(g(x^(2)))`
  3. `2xg^(′)(x^(2))e^(g(x^(2))`
  4. `2xg^(′)(2x)e^(g(x^(2)))`
Show Answers Only

`C`

Show Worked Solution

`f(x)=e^(g(x^2))`

`text{Using the chain rule (twice):}`

`f^(′)(x)` `=d/dx[g(x^2)] * e^(g(x^2))`
  `=2x*g^(′)(x^2)*e^(g(x^2))`

 
`=> C`

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

Calculus, 2ADV C2 EQ-Bank 11

Differentiate  `y = 2e^(−3x)` with respect to `x`.   (2 mark)

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`-6e^(-3x)`

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`y` `=2e^(-3x)`
`dy/dx` `=-3 xx 2e^(-3x)=-6e^(-3x)`

Filed Under: L&E Differentiation, Logs and Exponentials Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

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))`

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

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Trig Differentiation, Trigonometric Functions Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-7128-70-Trig Overlap, smc-7129-30-Tan, smc-7129-60-Chain Rule, smc-7129-70-Log/Exp Overlap, smc-965-10-Differentiation (base e), smc-965-50-Trig overlap, smc-967-10-Exponentials (base e), smc-967-50-Chain Rule, smc-967-80-Trig Overlap, smc-968-30-Tan, smc-968-60-Chain Rule, smc-968-70-Log/Exp Overlap

Calculus, 2ADV C2 EQ-Bank 4 MC

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

  1. `2/(x^(2x)ln2`
  2. `2/ln2 + 2log_2x`
  3. `log_2x+2/ln2`
  4. `2/ln2 xx log_2(x^(2x-1))`
Show Answers Only

`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`

Filed Under: L&E Differentiation, Log Calculus (Y12), Logs and Exponentials Tagged With: Band 4, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-7128-60-Log Laws required, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule, smc-967-60-New Reference Sheet, smc-967-70-Log Laws required

Calculus, 2ADV C2 EQ-Bank 13

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)`

 
`:. \ text{when}\ x=5,\ \ dy/dx=log_e 5 +2`

Filed Under: L&E Differentiation, Log Calculus (Y12), Logs and Exponentials Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule

Calculus, 2ADV C2 EQ-Bank 12

Let  `f(x) = (e^x)/((x^2-3))`.

Find `f^{′}(x)`.   (2 marks)

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`{e^x(x^2-2x-3)}/{(x^2-3)^2}`

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`text(Let) \ \ u = e^x \ \ => \ \ u^{′} = e^x`

 `v = (x^2-3) \ \ => \ \ v^{′}= 2x`

`f′(x)` `= {e^x(x^2-3)-2x e^x}/{(x^2-3)^2}`
  `= {e^x(x^2-2x-3)}/{(x^2-3)^2}`

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-40-Quotient Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-40-Quotient Rule

Calculus, 2ADV C2 2019 MET1 1a

Let  `y = (2e^(2x)-1)/e^x`.

Find  `(dy)/(dx)`.   (2 marks)

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`(dy)/(dx) = 2e^x + e^(-x)`

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`text(Method 1)`

`y` `= 2e^x-e^(-x)`
`(dy)/(dx)` `= 2e^x + e^(-x)`

  
`text(Method 2)`

`(dy)/(dx)` `= (4e^(2x) ⋅ e^x-(2e^(2x)-1) e^x)/(e^x)^2`
  `= (4e^(3x)-2e^(3x) + e^x)/e^(2x) `
  `= (2e^(2x) + 1)/e^x`

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-40-Quotient Rule, smc-7128-50-Chain Rule, smc-7128-60-Log Laws required, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-40-Quotient Rule, smc-967-50-Chain Rule, smc-967-70-Log Laws required

Calculus, 2ADV C2 EQ-Bank 32

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`

Filed Under: L&E Differentiation, Log Calculus (Y12), Logs and Exponentials Tagged With: Band 4, smc-7128-30-\(\log_e x\), smc-7128-60-Log Laws required, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-70-Log Laws required

Calculus, 2ADV C2 EQ-Bank 28

Differentiate  `5^(x^2)5x`.   (2 marks)

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`5^(x^2 + 1)(ln5*2x^2 + 1)`

Show Worked Solution
`y` `= 5^(x^2) * 5x`
`(dy)/(dx)` `= ln5*2x*5^(x^2)*5x + 5^(x^2)*5`
  `=5^(x^2)(ln5*10x^2 + 5)`
  `=5^(x^2 + 1)(ln5*2x^2 + 1)`

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-20-\(\large a^x\), smc-7128-35-Product Rule, smc-7128-50-Chain Rule, smc-965-20-Differentiation (base a), smc-967-15-Exponentials (base a), smc-967-30-Product Rule, smc-967-50-Chain Rule, smc-967-60-New Reference Sheet

Calculus, 2ADV C2 EQ-Bank 25

Differentiate  `3x  6^x`.   (2 marks)

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`3*6^x(xln6 +1)`

Show Worked Solution

`y= 3x * 6^x`

`text{Using the product rule:}`

`(dy)/(dx)= 3*6^x + ln6 * 6^x *3x= 3*6^x(1 + xln6)`

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-20-\(\large a^x\), smc-7128-35-Product Rule, smc-965-20-Differentiation (base a), smc-967-15-Exponentials (base a), smc-967-30-Product Rule, smc-967-60-New Reference Sheet

Calculus, 2ADV C2 EQ-Bank 26

Differentiate with respect to `x`:

`10^(5x^2-3x)`.   (2 marks)

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`(dy)/(dx) = ln 10  (10x-3) * 10^(5x^2-3x)`

Show Worked Solution

`y = 10^(5x^2-3x)`

`(dy)/(dx) = ln 10  (10x-3) * 10^(5x^2-3x)`

Filed Under: Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials Tagged With: Band 4, smc-7128-20-\(\large a^x\), smc-7128-50-Chain Rule, smc-965-20-Differentiation (base a), smc-967-15-Exponentials (base a), smc-967-50-Chain Rule, smc-967-60-New Reference Sheet

Calculus, 2ADV C2 EQ-Bank 15

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

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`2/(xln2)`

Show Worked Solution
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)`

Filed Under: L&E Differentiation, Log Calculus (Y12), Logs and Exponentials Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-964-10-Differentiation, smc-967-20-Logs, smc-967-60-New Reference Sheet

Calculus, 2ADV C2 2018 HSC 11g

Differentiate  `e^x/(x + 1)`.   (2 marks)

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`(xe^x)/(x + 1)^2`

Show Worked Solution

`y = e^x/(x + 1)`

`text(Differentiate using quotient rule:)`

`u = e^x\ \ \ \ \ \ \ v = x + 1`

`u^{\ prime } = e^x\ \ \ \ \ v^{\ prime }= 1`

`(dy)/(dx)` `= (u^{\ prime } v-u v ^{\ prime })/v^2`
  `= (e^x(x + 1)-e^x ⋅ 1)/(x + 1)^2`
  `= (x e^x)/(x + 1)^2`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-40-Quotient Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-40-Quotient Rule

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, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Trig Differentiation, Trigonometric Functions Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-50-Chain Rule, smc-7128-70-Trig Overlap, smc-7129-10-Sin, smc-7129-60-Chain Rule, 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 C2 2017 HSC 11d

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

Show Answers Only

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

Show Worked Solution

`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)`

Filed Under: L&E Differentiation, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule

Calculus, 2ADV C2 2017 HSC 3 MC

What is the derivative of  `e^(x^2)`?

  1. `x^2e^(x^2-1)`
  2. `2xe^(2x)`
  3. `2xe^(x^2)`
  4. `2e^(x^2)`
Show Answers Only

`C`

Show Worked Solution
`y` `= e^(x^2)`
`(dy)/(dx)` `= 2x  e^(x^2)`

`=>  C`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

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`
Show Answers Only

`B`

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

  
`=>  B`

Filed Under: Differentiation and Integration, L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation, Trig Differentiation, Trigonometric Functions Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-70-Trig Overlap, smc-7129-20-Cos, smc-7129-70-Log/Exp Overlap, smc-967-20-Logs, smc-967-80-Trig Overlap, smc-968-20-Cos, smc-968-70-Log/Exp Overlap

Calculus, 2ADV C2 2015 HSC 11f

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

Show Answers Only

`ln\x + 4/x +1`

Show Worked Solution

`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`

Filed Under: L&E Differentiation, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule

Calculus, 2ADV C2 2015 HSC 11e

Differentiate  `(e^x + x)^5`.   (2 marks)

Show Answers Only

`5 (e^x + 1) (e^x + x)^4`

Show Worked Solution
`y` `= (e^x + x)^5`
`(dy)/(dx)` `= 5 (e^x + x)^4 xx d/(dx) (e^x + x)`
  `= 5 (e^x + x)^4 xx (e^x + 1)`
  `= 5 (e^x + 1) (e^x + x)^4`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

Calculus, 2ADV C2 2007 HSC 2ai

Differentiate with respect to `x`:

`(2x)/(e^x + 1)`   (2 marks)

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`{2(e^x + 1-xe^x)}/(e^x + 1)^2`

Show Worked Solution

`y = (2x)/(e^x + 1)`

`u= 2x\ \ \ \ \v= e^x + 1`

`u ^{\ prime}=2\ \ \ \ \ v^{\ prime}= e^x`

`(dy)/(dx)` `= (u^{\ prime} v-uv^{\ prime})/v^2`
  `= {2(e^x + 1)- 2x(e^x)}/(e^x + 1)^2`
  `= (2e^x + 2-2x * e^x)/(e^x + 1)^2`
  `= {2(e^x + 1-xe^x)}/(e^x + 1)^2`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-40-Quotient Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-40-Quotient Rule

Calculus, 2ADV C3 2005 HSC 2d

Find the equation of the tangent to  `y = log_ex`  at the point  `(e, 1)`.   (2 marks)

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`y = x/e`

Show Worked Solution

`y = log_ex`

`dy/dx = 1/x`

`text(At)\ \ (e\ ,\ 1),\ \ m = 1/e`
 

`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e\ ,\ 1)`

`y-y_1` `= m(x-x_1)`
`y-1`  `= 1/e(x-e)`
`y-1`  `= x/e-1`
`y`  `= x/e`

Filed Under: Applied Calculus (L&E), L&E Differentiation, Logs and Exponentials, Tangents, Tangents and Normals Tagged With: Band 3, smc-1090-10-Find tangent given curve, smc-1090-50-Log/Exp Function, smc-7128-30-\(\log_e x\), smc-967-20-Logs

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`

Filed Under: L&E Differentiation, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule

Calculus, 2ADV C2 2009 HSC 2aii

Differentiate with respect to `x`.

`(e^x+1)^2`.   (2 marks) 

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`2e^x(e^x+1)`

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`y` `=(e^x+1)^2`
`dy/dx` `=2(e^x+1)^1xxd/(dx) (e^x+1)`
  `=2e^x(e^x+1)`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

Calculus, 2ADV C2 2011 HSC 1d

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

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`5/(5x+2)`

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

Filed Under: L&E Differentiation, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-964-10-Differentiation, smc-967-20-Logs

Calculus, 2ADV C2 2012 HSC 11d

Differentiate    `(3+e^(2x))^5`.   (2 marks) 

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`10e^(2x)(3+e^(2x))^4`

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`y=(3+e^(2x))^5`

`(dy)/dx` `=5(3+e^(2x))^4 xx  d/(dx)(3+e^(2x))`
  `=5(3+e^(2x))^4 xx 2e^(2x)`
  `=10e^(2x)(3+e^(2x))^4`

 

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-50-Chain Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-50-Chain Rule

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`

Filed Under: L&E Differentiation, Log Calculus, Log Calculus (Y12), Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-30-\(\log_e x\), smc-7128-35-Product Rule, smc-964-10-Differentiation, smc-967-20-Logs, smc-967-30-Product Rule

Calculus, 2ADV C2 2013 HSC 11d

Differentiate  `x^2e^x`    (2 marks)

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

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 `xe^x(x+2)`

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`text{Using the product rule}`

`text(Let)\ \ u=x^2,\ \ \ \ \ \ u^{\ prime}=2x`

`text(Let)\ \ v=e^x,\ \ \ \ \ \ v^{\ prime}=e^x`
  

`{d(uv)}/dx` `=u^{\ prime} v+v^{\ prime} u`
  `=2x e^x +x^2 e^x `
  `=xe^x(x+2)`

Filed Under: Exponential Calculus, Exponential Calculus (Y12), L&E Differentiation, Logs and Exponentials, Logs and Exponentials - Differentiation Tagged With: Band 3, smc-7128-10-\(\large e^x\), smc-7128-35-Product Rule, smc-965-10-Differentiation (base e), smc-967-10-Exponentials (base e), smc-967-30-Product Rule

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