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The slope of the tangent to the hyperbola \(2 x^2-3 y^2=6\) at \((3,2)\) is
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Differentiating the given equation of the curve
\(\begin{aligned}
& 4 x-6 y \cdot(\mathrm{dy} / \mathrm{dx})=0 \quad \therefore \mathrm{dy} / \mathrm{dx}=2 \mathrm{x} / 3 \mathrm{y} \\
& \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2}{3} \cdot \frac{3}{2}=1
\end{aligned}\)
\(\begin{aligned}
& 4 x-6 y \cdot(\mathrm{dy} / \mathrm{dx})=0 \quad \therefore \mathrm{dy} / \mathrm{dx}=2 \mathrm{x} / 3 \mathrm{y} \\
& \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2}{3} \cdot \frac{3}{2}=1
\end{aligned}\)
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