If 2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0,then th | vedclass

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Question

(B) Given the equation: $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) + \frac{d}{dx}(3xy)

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$-1$

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(B) Given the equation: $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) + \frac{d}{dx}(3xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(5y) + \frac{d}{dx}(6) = 0$. $4x + 3(x \frac{dy}{dx} + y) - 2y \frac{dy}{dx} + 4 - 5 \frac{dy}{dx} = 0$. Grouping the $\frac{dy}{dx}$ terms: $(3x - 2y - 5) \frac{dy}{dx} = -(4x + 3y + 4)$. $\frac{dy}{dx} = -\frac{4x + 3y + 4}{3x - 2y - 5}$. Substituting $(x, y) = (1, -2)$: $\frac{dy}{dx} = -\frac{4(1) + 3(-2) + 4}{3(1) - 2(-2) - 5} = -\frac{4 - 6 + 4}{3 + 4 - 5} = -\frac{2}{2} = -1$.

If $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0$,then the value of $\frac{dy}{dx}$ at $(x, y) = (1, -2)$ is

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