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Given $y = 2x(3x - 1)^5$, (a) find \( \frac{dy}{dx} \), giving your answer as a single fully factorised expression - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 5

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Given--$y-=-2x(3x---1)^5$,---(a)-find-\(-\frac{dy}{dx}-\),-giving-your-answer-as-a-single-fully-factorised-expression-Edexcel-A-Level Maths Pure-Question 3-2018-Paper 5.png

Given $y = 2x(3x - 1)^5$, (a) find \( \frac{dy}{dx} \), giving your answer as a single fully factorised expression. (b) Hence find the set of values of \( x \) ... show full transcript

Worked Solution & Example Answer:Given $y = 2x(3x - 1)^5$, (a) find \( \frac{dy}{dx} \), giving your answer as a single fully factorised expression - Edexcel - A-Level Maths Pure - Question 3 - 2018 - Paper 5

Step 1

find \( \frac{dy}{dx} \)

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Answer

To find ( \frac{dy}{dx} ), we will apply the product rule of differentiation. Let ( u = 2x ) and ( v = (3x - 1)^5 ). The product rule states that:

dydx=uv+uv\frac{dy}{dx} = u'v + uv'

First, we compute ( u' ) and ( v' ):

  • ( u' = 2 )
  • For ( v' ), we use the chain rule:
    ( v = (3x - 1)^5 ), so applying the chain rule: ( v' = 5(3x - 1)^4 \cdot 3 = 15(3x - 1)^4 )

Now substituting back into the product rule:

dydx=2(3x1)5+2x(15(3x1)4)\frac{dy}{dx} = 2(3x - 1)^5 + 2x(15(3x - 1)^4)

Factoring out the common term ( (3x - 1)^4 ):

dydx=(3x1)4[2(3x1)+30x]\frac{dy}{dx} = (3x - 1)^4 \left[ 2(3x - 1) + 30x \right]

This simplifies to:

dydx=(3x1)4(32x2)\frac{dy}{dx} = (3x - 1)^4 (32x - 2)

Thus, the final answer is:

dydx=2(3x1)4(16x1)\frac{dy}{dx} = 2(3x - 1)^4 (16x - 1)

Step 2

Hence find the set of values of \( x \) for which \( \frac{dy}{dx} \leq 0 \)

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Answer

To find the set of values for which ( \frac{dy}{dx} \leq 0 ), we analyze the expression:

dydx=2(3x1)4(16x1)\frac{dy}{dx} = 2(3x - 1)^4 (16x - 1)

The factor ( 2(3x - 1)^4 ) is always non-negative since it is raised to an even power and multiplied by a positive constant. Thus, the sign of ( \frac{dy}{dx} ) is determined solely by ( 16x - 1 ).

Setting ( 16x - 1 \leq 0 ) gives:

16x1x11616x \leq 1 \Rightarrow x \leq \frac{1}{16}

Thus, the set of values of ( x ) for which ( \frac{dy}{dx} \leq 0 ) is:

x(,116]x \in \left( -\infty, \frac{1}{16} \right]

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