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

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Question

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

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) Given the equation: $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(4y^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(3y) + \frac{d}{dx}(4) = 0$. $4x - 3(y + x \frac{dy}{dx}) + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$. $4x - 3y - 3x \frac{dy}{dx} + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$. Now, substitute the point $(3, 2)$ where $x = 3$ and $y = 2$: $4(3) - 3(2) - 3(3) \frac{dy}{dx} + 8(2) \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$. $12 - 6 - 9 \frac{dy}{dx} + 16 \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$. $8 + 4 \frac{dy}{dx} = 0$. $4 \frac{dy}{dx} = -8$. $\frac{dy}{dx} = -2$.

If $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$, then $\left(\frac{dy}{dx}\right)_{(3,2)} = $

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