TS EAMCET · Maths · Differentiation
If \(x^2+y^2=t-\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}\) then \(\frac{d y}{d x}=\)
- A \(\frac{x}{y}\)
- B \(\frac{-x}{y}\)
- C \(\frac{y}{x}\)
- D \(\frac{-y}{x}\)
Answer & Solution
Correct Answer
(D) \(\frac{-y}{x}\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{t}-\frac{1}{\mathrm{t}}\) \(\begin{aligned} & \text { and } x^4+y^4=t^2+\frac{1}{t^2} \\ & \Rightarrow x^4+y^4=x^4+y^4+2 x^2 y^2+2 \\ & \Rightarrow x^2 y^2+1=0 \end{aligned}\) Now \(\frac{d y}{d x} \Rightarrow 2 x y^2+x^2 2 y y l=0\)…
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