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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. (b) Hence find the set of values of $x$ for w... 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 \( \frac{dy}{dx} \)
Answer
To find ( \frac{dy}{dx} ), we use the product rule. Let:
- ( u = 2x )
- ( v = (3x - 1)^5 )
Then: [ \frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx} ]
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Calculate ( \frac{du}{dx} ): [ \frac{du}{dx} = 2 ]
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Calculate ( \frac{dv}{dx} ) using the chain rule: [ \frac{dv}{dx} = 5(3x - 1)^4 \cdot 3 = 15(3x - 1)^4 ]
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Substitute into the product rule: [ \frac{dy}{dx} = 2(3x - 1)^5 + 2x \cdot 15(3x - 1)^4 ]
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Factor out common terms: [ \frac{dy}{dx} = 2(3x - 1)^4 \left( (3x - 1) + 15x \right) ]
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Simplify: [ \frac{dy}{dx} = 2(3x - 1)^4 (18x - 1) ]
Step 2
Hence find the set of values of $x$ for which \( \frac{dy}{dx} \leq 0 \)
Answer
From the expression ( \frac{dy}{dx} = 2(3x - 1)^4 (18x - 1) ), we analyze when this is less than or equal to zero.
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Since ( 2(3x - 1)^4 ) is always positive (it is a square term), we only need to check: [ 18x - 1 \leq 0 ]
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Solve for ( x ): [ 18x \leq 1 \implies x \leq \frac{1}{18} ]
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Thus, the solution set is: [ x \in \left( -\infty, \frac{1}{18} \right] ]