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Differentiate with respect to $x$, giving each answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1
Question 9
Differentiate with respect to $x$, giving each answer in its simplest form. (a) $(1 - 2x)^2$ (b) \( \frac{x^3 + 6 \sqrt{x}}{2x^2} \)
Worked Solution & Example Answer:Differentiate with respect to $x$, giving each answer in its simplest form - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1
Step 1
(a) \((1 - 2x)^2")
Answer
To differentiate ((1 - 2x)^2") with respect to , we will use the chain rule:
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First, let ( u = 1 - 2x ). Then, ( \frac{du}{dx} = -2 ).
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Use the formula for differentiation: ( \frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} ).
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Substitute back: [ \frac{d}{dx}((1 - 2x)^2) = 2(1 - 2x)(-2) = -4(1 - 2x) ]
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Expanding this gives: [ -4 + 8x ] Hence, the answer is (-4 + 8x).
Step 2
(b) \( \frac{x^3 + 6 \sqrt{x}}{2x^2} \)
Answer
To differentiate ( \frac{x^3 + 6 \sqrt{x}}{2x^2} ), we will use the quotient rule, which states:
[ \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} ]
Where:
- ( f(x) = x^3 + 6 \sqrt{x} )
- ( g(x) = 2x^2 )
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Differentiate ( f(x) ): [ f'(x) = 3x^2 + 3x^{-\frac{1}{2}} ] Simplifying gives: ( f'(x) = 3x^2 + \frac{3}{\sqrt{x}} )
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Differentiate ( g(x) ): [ g'(x) = 4x ]
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Apply the quotient rule: [ \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{(3x^2 + 3x^{-\frac{1}{2}})(2x^2) - (x^3 + 6\sqrt{x})(4x)}{(2x^2)^2}\n ]
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Simplify the numerator:
- First term: ( (3x^2 + \frac{3}{\sqrt{x}})(2x^2) = 6x^4 + 6x^{\frac{3}{2}} )
- Second term: ( (x^3 + 6\sqrt{x})(4x) = 4x^4 + 24x^{\frac{3}{2}} )
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Combine: [ (6x^4 + 6x^{\frac{3}{2}} - 4x^4 - 24x^{\frac{3}{2}}) = 2x^4 - 18x^{\frac{3}{2}} ]
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Final answer: [ \frac{2x^4 - 18x^{\frac{3}{2}}}{4x^4} \ = \frac{x^4 - 9x^{\frac{3}{2}}}{2x^4}``
In its simplest form, this is ( \frac{1 - \frac{9}{x^{\frac{1}{2}}}}{2} ).