by differentiating, $\displaystyle x^2 + xy - 2y^2 = 0$ show that $\displaystyle \frac{dy}{dx} = 1 $ or $\displaystyle \frac{-1}{2}$ Thanks
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Implicit differentiation yields: $\displaystyle 2x + xy' + y - 4yy' = 0$ $\displaystyle xy' - 4yy' = -2x - y$ $\displaystyle y'(x - 4y) = -2x - y$ $\displaystyle y' = \frac{-2x - y}{x - 4y}$ That's as far as I can go without knowing more information.
that is the same with what I got.
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