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Calculus, 2ADV C1 EO-Bank 14 v1

Evaluate `f^{′}(1)`, where `f(x) = x^2 / sqrt(2x + 3)`. (4 marks)

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`9 / (5sqrt5)`

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`f(x) = x^2(2x + 3)^(-1/2)`

`f^{′}(x)` `= 2x(2x + 3)^(-1/2) + x^2(-1/2)(2x + 3)^(-3/2)(2)`

`= (2x)/(sqrt(2x + 3)) – (x^2)/(2x + 3)^(3/2)`

`= [2x(2x + 3) – x^2] / (2x + 3)^(3/2)`

`= (3x^2 + 6x) / (2x + 3)^(3/2)`

`f^{′}(1)` `= (3(1)^2 + 6(1)) / (2(1) + 3)^(3/2)`

`= 9 / (5sqrt5)`

Calculus, 2ADV C1 EO-Bank 7

Differentiate  `2x(1-4x)^5`  with respect to `x`.   (2 marks)

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`y^{′}=2(1-4x)^4(1-24x)`

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`y=2x(1-4x)^5`

`text{Using the product and chain rules:}`

`y^{′}` `=2 xx (1-4x)^5-40x(1-4x)^4`  
  `=2(1-4x)^4(1-4x-20x)`  
  `=2(1-4x)^4(1-24x)`  

Calculus, 2ADV C1 2023 MET2 11 MC

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-2)=-7,\ g(-2)=8\)  and  \(f^{′}(-2)=3,\ g^{′}(-2)=2\).

The gradient of the graph  \(y=f(x)\times g(x)\)  at the point where  \(x=-2\)  is

  1. \(-6\)
  2. \(0\)
  3. \(6\)
  4. \(10\)
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\(D\)

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\(\text{Using the Product Rule when}\ \ x=-2:\)

\(\dfrac{d}{dx}(f(x)\times g(x))\) \(=f(x)g^{′}(x)+g(x)f^{′}(x)\)
  \(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\)
  \(=-7\times 2+8\times 3\)
  \(=10\)

 
\(\Rightarrow D\)


♦♦♦ Mean mark 22%.