Find frac{dy}{dx} for the equation xy + y^2 = | vedclass

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(A) The given equation is $xy + y^2 = 	an x + y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(	an

Vedclass · Accepted Answer

$\frac{\sec^2 x - y}{x + 2y - 1}$

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(A) The given equation is $xy + y^2 = 	an x + y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(	an x + y)$ Using the product rule for $xy$ and the chain rule for $y^2$: $y \cdot \frac{d}{dx}(x) + x \cdot \frac{dy}{dx} + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$ $y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$ Grouping the terms containing $\frac{dy}{dx}$ on one side: $x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y$ $(x + 2y - 1) \frac{dy}{dx} = \sec^2 x - y$ Therefore,$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$.

Find $\frac{dy}{dx}$ for the equation $xy + y^2 = \tan x + y$.

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