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 \(\frac{y}{x}\)
- B \(\frac{y^2}{x^2}\)
- C \(\sqrt{\frac{y}{x}}\)
- D \(-\frac{y}{x}\)
Answer & Solution
Correct Answer
(D) \(-\frac{y}{x}\)
Step-by-step Solution
Detailed explanation
\((x^2+y^2)^2 = (t-\frac{1}{t})^2 = t^2-2+\frac{1}{t^2}\) \((x^2+y^2)^2 = (x^4+y^4)-2\) \(x^4+2x^2y^2+y^4 = x^4+y^4-2\) \(2x^2y^2 = -2\) \(x^2y^2 = -1\) \(\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(-1)\) \(2xy^2+x^2(2y\frac{dy}{dx}) = 0\) \(2xy^2+2x^2y\frac{dy}{dx} = 0\)…
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