If sin (xy) + frac{x}{y} = {x^2} - y, then fr | vedclass

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Question

(A) Given the equation: $\sin(xy) + \frac{x}{y} = x^2 - y$. Differentiating both sides with respect to $x$: $\cos(xy) \cdot (y + x \frac{dy}{dx}) + \f

Vedclass · Accepted Answer

$\frac{y(2xy^2 - y - y^3\cos(xy))}{x(y^2\cos(xy) - x + y^2)}$

Vedclass · Answer

(A) Given the equation: $\sin(xy) + \frac{x}{y} = x^2 - y$. Differentiating both sides with respect to $x$: $\cos(xy) \cdot (y + x \frac{dy}{dx}) + \frac{y(1) - x \frac{dy}{dx}}{y^2} = 2x - \frac{dy}{dx}$. Multiply by $y^2$ to simplify: $y^2 \cos(xy) (y + x \frac{dy}{dx}) + y - x \frac{dy}{dx} = 2xy^2 - y^2 \frac{dy}{dx}$. Expand and group the $\frac{dy}{dx}$ terms: $y^3 \cos(xy) + xy^2 \cos(xy) \frac{dy}{dx} + y - x \frac{dy}{dx} = 2xy^2 - y^2 \frac{dy}{dx}$. $\frac{dy}{dx} (xy^2 \cos(xy) - x + y^2) = 2xy^2 - y - y^3 \cos(xy)$. Therefore,$\frac{dy}{dx} = \frac{2xy^2 - y - y^3 \cos(xy)}{xy^2 \cos(xy) - x + y^2}$. Factoring out $y$ from the numerator: $\frac{dy}{dx} = \frac{y(2xy - 1 - y^2 \cos(xy))}{xy^2 \cos(xy) - x + y^2}$.

If $\sin (xy) + \frac{x}{y} = {x^2} - y,$ then $\frac{dy}{dx} = $

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