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Calculus, MET1 2023 VCAA 1b

Let  \(f(x)=\sin(x)e^{2x}\).

Find  \(f^{'}\Big(\dfrac{\pi}{4}\Big)\).   (2 marks)

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\(\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \text{or}\ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

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

\(f'(x)\) \(=e^{2x}\cos(x)+2e^{2x}\sin(x)\)
  \(=e^{2x}\Big(\cos(x)+2\ \sin(x)\Big)\)
\(\therefore\ f’\Big(\dfrac{\pi}{4}\Big)\) \(=e^{2(\frac{\pi}{4})}\Bigg(\cos(\dfrac{\pi}{4})+2\ \sin(\dfrac{\pi}{4})\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1}{\sqrt{2}}+\sqrt{2}\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1+\sqrt2 \times \sqrt2}{\sqrt2} \Bigg) \)
  \(=\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \ \text{or}\ \ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

Calculus, MET1 2008 VCAA 1b

Let  `f(x) = xe^(3x)`.  Evaluate  `f^{prime}(0)`.   (3 marks)

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`1`

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`text(Using Product Rule:)`

`(gh)^{prime} = g^{prime}h + gh^{prime}`

`f^{prime}(x)` `= x(3e^(3x)) + 1 xx e^(3x)`
`:.f^{prime}(0)` `= 0 + e^0`
  `= 1`

Calculus, MET1 2016 VCAA 1b

Let  `f(x) = x^2e^(5x)`.

Evaluate  `f^{\prime}(1)`.   (2 marks)

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`7e^5`

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`text(Using Product Rule:)`

`(fg)^{\prime}` `= f^{\prime}g + fg^{\prime}`
`f^{′}(x)` `= 2xe^(5x) + 5x^2 e^(5x)`
`f^{′}(1)` `= 2(1)e^(5(1)) + 5(1)^2 e^(5(1))`
  `= 7e^5`

Calculus, MET2 2009 VCAA 7 MC

For  `y = e^(2x) cos (3x)`  the rate of change of `y` with respect to `x` when  `x = 0`  is

  1. `0`
  2. `2`
  3. `3`
  4. `– 6`
  5. `– 1`
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`B`

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`y` `= e^(2x) cos (3x)`
`dy/dx` `=e^(2x) xx -3sin(3x) + 2e^(2x) xx cos (3x)`
  `=e^(2x)(-3sin(3x) + 2cos(3x))`

 
`text(When)\ \ x = 0,`

`dy/dx= 2`

`=>   B`

Calculus, MET1 2010 VCAA 1a

Differentiate  `x^3 e^(2x)`  with respect to `x`.   (2 marks)

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

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    `text(Using Product rule:)`

`(fg)^{prime}` `= f^{prime}g + fg^{prime}`
`d/(dx) (x^3 e^(2x))` `= 3x^2 e^(2x) + 2x^3 e^(2x)`