COMEDK · Maths · 26. Differentiation
\(\text { If } x^2+y^2=t+\dfrac{1}{t} \text { and } x^4+y^4=t^2+\dfrac{1}{t^2} \text { then } \dfrac{d y}{d x}=\)
- A \(\dfrac{x}{2 y}\)
- B \(\dfrac{y}{x}\)
- C \(-\dfrac{y}{x}\)
- D \(-\dfrac{x}{2 y}\)
Answer & Solution
Correct Answer
(C) \(-\dfrac{y}{x}\)
Step-by-step Solution
Detailed explanation
Given \(x^2 + y^2 = t + \dfrac{1}{t}\) and \(x^4 + y^4 = t^2 + \dfrac{1}{t^2}\). Squaring the first equation: \((x^2 + y^2)^2 = (t + \dfrac{1}{t})^2\) \(x^4 + y^4 + 2x^2y^2 = t^2 + \dfrac{1}{t^2} + 2\) Substituting \(x^4 + y^4 = t^2 + \dfrac{1}{t^2}\) into the expanded equation:…
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