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If \(x^2+y^2=t-\frac{1}{t}\) and \(x^4+y^4=t^2+\frac{1}{t^2}\), then \(\frac{d y}{d x}=\)
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Verified Answer
The correct answer is:
\(\frac{1}{x^3 y}\)
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
\(\begin{aligned}
x^4+y^4+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \\
t^2+\frac{1}{t^2}+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \quad[\because \text {from Eq. (i)}] \\
x^2 y^2 & =-1 \\
y^2 & =\frac{-1}{x^2}
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
2 y y^{\prime} & =\frac{2}{x^3} \\
y^{\prime} & =\frac{1}{x^3 y}
\end{aligned}\)
\(\therefore\) Hence, answer is (d).
\(\begin{aligned}
& x^4+y^4=t^2+\frac{1}{t^2} \quad \ldots (i) \\
& x^2+y^2=t-\frac{1}{t}
\end{aligned}\)
On Squaring both sides, we get
\(\begin{aligned}
x^4+y^4+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \\
t^2+\frac{1}{t^2}+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \quad[\because \text {from Eq. (i)}] \\
x^2 y^2 & =-1 \\
y^2 & =\frac{-1}{x^2}
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
2 y y^{\prime} & =\frac{2}{x^3} \\
y^{\prime} & =\frac{1}{x^3 y}
\end{aligned}\)
\(\therefore\) Hence, answer is (d).
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