If 3y^2 - 2xy - x = 0, then the value of frac | vedclass

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Question

(C) Given equation: $3y^2 - 2xy - x = 0$.First, find the value of $x$ when $y = 2$: $3(2)^2 - 2x(2) - x = 0 \implies 12 - 4x - x = 0 \implies 12 = 5x

Vedclass · Accepted Answer

$\frac{25}{36}$

Vedclass · Answer

(C) Given equation: $3y^2 - 2xy - x = 0$.First, find the value of $x$ when $y = 2$: $3(2)^2 - 2x(2) - x = 0 \implies 12 - 4x - x = 0 \implies 12 = 5x \implies x = \frac{12}{5}$.Differentiate the equation with respect to $x$: $\frac{d}{dx}(3y^2) - \frac{d}{dx}(2xy) - \frac{d}{dx}(x) = 0$.$6y \frac{dy}{dx} - (2y + 2x \frac{dy}{dx}) - 1 = 0$.Substitute $y = 2$ and $x = \frac{12}{5}$: $6(2) \frac{dy}{dx} - 2(2) - 2(\frac{12}{5}) \frac{dy}{dx} - 1 = 0$.$12 \frac{dy}{dx} - 4 - \frac{24}{5} \frac{dy}{dx} - 1 = 0$.$(12 - \frac{24}{5}) \frac{dy}{dx} = 5$.$(\frac{60 - 24}{5}) \frac{dy}{dx} = 5$.$\frac{36}{5} \frac{dy}{dx} = 5$.$\frac{dy}{dx} = \frac{25}{36}$.

If $3y^2 - 2xy - x = 0$, then the value of $\frac{dy}{dx}$ at $y = 2$ is...

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