Find frac{dy}{dx} if 2x^2 - 3xy + y^2 + x + 2 | vedclass

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Question

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \

Vedclass · Accepted Answer

$\frac{3y - 4x - 1}{2y - 3x + 2}$

Vedclass · Answer

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$ $4x - 3(y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$ $4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$ Grouping the $\frac{dy}{dx}$ terms: $(2y - 3x + 2) \frac{dy}{dx} = 3y - 4x - 1$ Therefore,$\frac{dy}{dx} = \frac{3y - 4x - 1}{2y - 3x + 2}$. Thus,option $A$ is correct.

Find $\frac{dy}{dx}$ if $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.

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