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If $x^2+y^2=t-\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}$, then $x^3 y \frac{d y}{d x}=$
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$\begin{aligned} & x^4+y^4=\left(t-\frac{1}{t}\right)^2+2=\left(x^2+y^2\right)^2+2 \\ & \Rightarrow x^2 y^2=-1 \Rightarrow y^2=-\frac{1}{x^2}\end{aligned}$
Differentiating, we get $2 y \frac{d y}{d x}=\frac{2}{x^3}$ or $x^3 y \frac{d y}{d x}=1$
Differentiating, we get $2 y \frac{d y}{d x}=\frac{2}{x^3}$ or $x^3 y \frac{d y}{d x}=1$
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