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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)))`
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`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 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 28

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

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

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`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 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 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 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)`
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`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 2015 HSC 11e

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

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`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 2009 HSC 2aii

Differentiate with respect to `x`.

`(e^x+1)^2`.   (2 marks) 

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

Show Worked Solutions
`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 2012 HSC 11d

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

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

Show Worked Solutions

`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

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