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If $x=\frac{1+\log t}{t^{2}}, y=\frac{3+2 \log t}{t}$, find $\frac{d y}{d x}$


$x=\frac{1+\log t}{t^{2}}$ and $y=\frac{3+2 \log t}{t}$

$\Rightarrow \frac{d x}{d t}=\frac{t-2 t-2 t \log t}{t^{4}}$ and $\frac{d y}{d t}=\frac{2-3-2 \log t}{t^{2}}$

$\Rightarrow \frac{d x}{d t}=\frac{-1-2 \log t}{t^{3}}$ and $\frac{d y}{d t}=\frac{-1-2 \log t}{t^{2}}$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{-1-2 \log t}{t^{2}}}{\frac{-1-2 \log t}{t^{3}}}=t$

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