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Calculus, 2ADV C1 EQ-Bank 7

Differentiate  `x(1+3x)^5`  with respect to `x`.   (2 marks)

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`y^{′}=(1+3x)^4(18 x+1)`

Show Worked Solution

`y=x(1+3x)^5`

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

`y^{′}` `=1 xx (1+3x)^5+15x(1+3x)^4`  
  `=(1+3x)^4(1+3x+15 x)`  
  `=(1+3x)^4(18 x+1)`  

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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\(D\)

Show Worked Solution

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

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

 
\(\Rightarrow D\)


♦♦♦ Mean mark 22%.