AP EAMCET · Maths · Differentiation
If \(x^2+y^2=t+\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}\), then \(x^3 y \frac{d y}{d x}\) is equal to
- A \(-1\)
- B \(1\)
- C \(0\)
- D \(t\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
Given, \(x^4+y^4=t^2+\frac{1}{t^2}\)…
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