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It is given that $y = (1 + x^2) ext{ln}(1 + x)$ - CIE - A-Level Further Maths - Question 5 - 2013 - Paper 1
Question 5
It is given that $y = (1 + x^2) ext{ln}(1 + x)$. Find \( \frac{d^3y}{dx^3} \). Prove by mathematical induction that, for every integer $n \geq 3$, \( \frac{d^ny}{d... show full transcript
Worked Solution & Example Answer:It is given that $y = (1 + x^2) ext{ln}(1 + x)$ - CIE - A-Level Further Maths - Question 5 - 2013 - Paper 1
Step 1
Find \( \frac{d^3y}{dx^3} \)
Answer
To find the third derivative of ( y = (1 + x^2) \text{ln}(1 + x) ), we will first differentiate ( y ) three times:
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First Derivative: Using the product rule, we have: [ \frac{dy}{dx} = (1+x^2) \cdot \frac{1}{1+x} + \text{ln}(1+x) \cdot 2x = \frac{1+x^2}{1+x} + 2x \text{ln}(1+x). ]
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Second Derivative: Differentiate again: [ \frac{d^2y}{dx^2} = \frac{d}{dx}\left( \frac{1+x^2}{1+x} + 2x \text{ln}(1+x) \right). ]
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Third Derivative: Differentiate again: [ \frac{d^3y}{dx^3} \text{ (using differentiation rules including the product and quotient rules)}. ]
Step 2
Prove by mathematical induction for every integer \( n \geq 3 \)
Answer
To use mathematical induction, we start with the base case and then show that if it holds for ( n = k ), it also holds for ( n = k + 1 ).
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Base Case: For ( n = 3 ): [ \frac{d^3y}{dx^3} = (-1)^{3} \frac{2(3-3)!}{(1+x)^{3-2}} = -2. ] We will calculate ( \frac{d^3y}{dx^3} ) using the derivatives acquired in the previous steps to ensure it equals -2.
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Inductive Hypothesis: Assume it holds for ( n = k ): [ \frac{d^{k}y}{dx^{k}} = (-1)^{k} \frac{2(k-3)!}{(1+x)^{k-2}}. ]
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Inductive Step: We need to show it holds for ( n = k + 1 ): Find ( \frac{d^{k+1}y}{dx^{k+1}} ) using the chain and product rules, which must yield: [ \frac{d^{k+1}y}{dx^{k+1}} = (-1)^{k+1} \frac{2((k+1)-3)!}{(1+x)^{(k+1)-2}}. ]
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Conclusion: By mathematical induction, it holds that ( \frac{d^ny}{dx^n} = (-1)^{n} \frac{2(n-3)!}{(1+x)^{n-2}} ) for all integers ( n \geq 3 ).