TS EAMCET · Maths · Differentiation
If \(x^2+y^2=t+\frac{2}{t}\) and \(x^4+y^4=t^2+\frac{4}{t^2}\), then \(x^3 y \frac{d y}{d x}\) equals
- A -1
- B -2
- C \(\frac{y}{x}\)
- D \(x y\)
Answer & Solution
Correct Answer
(B) -2
Step-by-step Solution
Detailed explanation
We have, \[ x^2+y^2=t+\frac{2}{t} \text { and } x^4+y^4=t^2+\frac{4}{t^2} \] Now, \[ x^2+y^2=t+\frac{2}{t} \] On squaring both sides, we get…
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