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If $x^2+y^2=t+\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}$, then $x^3 y \frac{d y}{d x}$ is equal to
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$-1$
Given,
$x^4+y^4=t^2+\frac{1}{t^2}$
$\begin{aligned} & =\left(t+\frac{1}{t}\right)^2-2 \\ & =\left(x^2+y^2\right)^2-2 \quad\left(\because x^2+y^2=t+\frac{1}{t}, \text { given }\right) \\ & \Rightarrow \quad x^4+y^4=x^4+y^4+2 x^2 y^2-2 \\ & \Rightarrow \quad x^2 y^2=1 \\ & \Rightarrow \quad y^2=\frac{1}{x^2}\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned} & 2 y \frac{d y}{d x}=-\frac{2}{x^3} \\ & x^3 y \frac{d y}{d x}=-1\end{aligned}$
$x^4+y^4=t^2+\frac{1}{t^2}$
$\begin{aligned} & =\left(t+\frac{1}{t}\right)^2-2 \\ & =\left(x^2+y^2\right)^2-2 \quad\left(\because x^2+y^2=t+\frac{1}{t}, \text { given }\right) \\ & \Rightarrow \quad x^4+y^4=x^4+y^4+2 x^2 y^2-2 \\ & \Rightarrow \quad x^2 y^2=1 \\ & \Rightarrow \quad y^2=\frac{1}{x^2}\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned} & 2 y \frac{d y}{d x}=-\frac{2}{x^3} \\ & x^3 y \frac{d y}{d x}=-1\end{aligned}$
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