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Calculus, MET2 2023 VCAA 7 MC

Let  \(f(x)=\log_{e}x\), where  \(x>0\)  and  \(g(x)=\sqrt{1-x}\), where  \(x<1\).

The domain of the derivative of \((f\circ g)(x)\) is

  1. \(x\in R\)
  2. \(x\in (-\infty, 1]\)
  3. \(x\in (-\infty, 1)\)
  4. \(x\in (0, \infty)\)
  5. \(x\in (0, 1)\)
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\(C\)

Show Worked Solution

\(\text{Given }f(x)=\log_{e}x\ \ \text{and}\ \ g(x)=\sqrt{1-x}\)

\((f\circ g)(x)=\log_{e}\sqrt{1-x}=\dfrac{1}{2}\log_{e}(1-x)\)

\((f\circ g)^{′}(x)=\dfrac{1}{2}\times\dfrac{-1}{1-x}=\dfrac{1}{2(x-1)}\ \text{  where}\ \ x<1\)

\(\Rightarrow C\)

Calculus, MET2 2020 VCAA 7 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)))`
  5. `2xg^(')(x^(2))e^(g(2x))`
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`