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5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1 + x^2} \) - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1
Question 25
5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1 + x^2} \)
Worked Solution & Example Answer:5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \( \frac{dy}{dx} = \frac{-1}{1 + x^2} \) - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1
Step 1
(a) y = x² ln 2x
Answer
To differentiate the function ( y = x^2 \ln 2x ), we need to apply the product rule. The product rule states that ( \frac{d}{dx}(uv) = u'v + uv' ) where ( u = x^2 ) and ( v = \ln 2x ).
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Differentiate ( u = x^2 ): [ u' = 2x ]
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Differentiate ( v = \ln 2x ): Using the chain rule: [ v' = \frac{1}{2x} \cdot \frac{d}{dx}(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} ]
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Now apply the product rule: [ \frac{dy}{dx} = u'v + uv' = 2x \ln 2x + x^2 \cdot \frac{1}{x} = 2x \ln 2x + x ]
Thus, the full answer for part (a) is: [ \frac{dy}{dx} = 2x \ln 2x + x ]
Step 2
(b) y = (x + sin 2x)³
Answer
To differentiate ( y = (x + \sin 2x)^3 ), we apply the chain rule:
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Let ( u = x + \sin 2x ), then ( y = u^3 ). So we differentiate as follows: [ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ]
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Differentiate ( y = u^3 ): [ \frac{dy}{du} = 3u^2 = 3(x + \sin 2x)^2 ]
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Now find ( \frac{du}{dx} ): [ \frac{du}{dx} = 1 + \cos 2x \cdot 2 = 1 + 2 \cos 2x ]
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Combine using the chain rule: [ \frac{dy}{dx} = 3(x + \sin 2x)^2 \cdot (1 + 2\cos 2x) ]
Therefore, the answer for part (b) is: [ \frac{dy}{dx} = 3(x + \sin 2x)^2 (1 + 2\cos 2x) ]
Step 3
(ii) Given that x = cot y, show that dy/dx = -1/(1 + x²)
Answer
Given ( x = \cot y ), we differentiate both sides with respect to x:
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Using the identity, we have: [ \frac{dx}{dy} = -\csc^2 y ]
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Now, we need to find ( \frac{dy}{dx} ): By using reciprocal relation: [ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = -\frac{1}{\csc^2 y} = -\sin^2 y ]
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Next, we use the identity for ( \cot y ) which gives us: [ \cot^2 y + 1 = \csc^2 y ] Therefore, substituting ( x = \cot y ) yields: [ \csc^2 y = 1 + x^2 ]
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This lets us express ( \frac{dy}{dx} ): [ \frac{dy}{dx} = -\sin^2 y = -\frac{1}{1+x^2} ]
We have now shown that: [ \frac{dy}{dx} = \frac{-1}{1 + x^2} ]