If \(x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}\) and \(x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is equal to

\(x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}\)
Squaring on both sides, we get
\(x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2\)
\(\therefore \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2 +\frac{1}{\mathrm{t}^2}+2 \)
\( \ldots\left[\because x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}, \text { given }\right]\)
\(\begin{aligned}
& 2 x^2 y^2=2 \\
& x^2 y^2=1
\end{aligned}\)
differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \\
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x}
\end{aligned}\)