The original equation: $x^2 - xy + y^2 = 3$
After Implicit Differentiation: $\dfrac{\mathrm dy}{\mathrm dx} = (-2x+y)(-x+2y)$
If I want to find $x$ and $y$ when $ \frac{\mathrm dy}{\mathrm dx} = 0$ how would I do that?
The original equation: $x^2 - xy + y^2 = 3$
After Implicit Differentiation: $\dfrac{\mathrm dy}{\mathrm dx} = (-2x+y)(-x+2y)$
If I want to find $x$ and $y$ when $ \frac{\mathrm dy}{\mathrm dx} = 0$ how would I do that?
Solve -2x + y = 0 and -x + 2y = 0. The values where the two equations equal zero is where dy/dx will equal zero as well. Although, as mentioned above, one or the other will be zero, and since anything multiplied by zero is zero, the whole equation will equal zero.