# How do you determine dy/dx given x^2+y^2=sqrt7?

We differentiate both sides with respect to x

$d \frac{{x}^{2} + {y}^{2}}{\mathrm{dx}} = d \frac{\sqrt{7}}{\mathrm{dx}}$

$2 \cdot x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$x + y \cdot \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

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