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Calculus, 2ADV C2 SM-Bank 10

Differentiate with respect to \(x\) :

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

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

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\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}

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

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

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

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))`
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`B`

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`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 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`
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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 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`

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`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`