AP EAMCET · Maths · Differentiation
If \(x^2+y^2=t-\frac{1}{t}\) and \(x^4+y^4=t^2+\frac{1}{t^2}\), then \(\frac{d y}{d x}=\)
- A \(\frac{2}{x^3}\)
- B \(\frac{2}{x^3 y}\)
- C \(\frac{1}{x^3}\)
- D \(\frac{1}{x^3 y}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{x^3 y}\)
Step-by-step Solution
Detailed explanation
Given, \(\begin{aligned} & x^4+y^4=t^2+\frac{1}{t^2} \quad \ldots (i) \\ & x^2+y^2=t-\frac{1}{t} \end{aligned}\) On Squaring both sides, we get…
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