If sqrt{x} + sqrt{y} = sqrt{xy},then frac{dy} | vedclass

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(A) Given the equation $\sqrt{x} + \sqrt{y} = \sqrt{xy}$.Dividing both sides by $\sqrt{xy}$,we get:$\frac{\sqrt{x}}{\sqrt{xy}} + \frac{\sqrt{y}}{\sqrt

Vedclass · Accepted Answer

$-\left(\frac{y}{x}\right)^{\frac{3}{2}}$

Vedclass · Answer

(A) Given the equation $\sqrt{x} + \sqrt{y} = \sqrt{xy}$.Dividing both sides by $\sqrt{xy}$,we get:$\frac{\sqrt{x}}{\sqrt{xy}} + \frac{\sqrt{y}}{\sqrt{xy}} = 1$$\frac{1}{\sqrt{y}} + \frac{1}{\sqrt{x}} = 1$$y^{-1/2} + x^{-1/2} = 1$Differentiating both sides with respect to $x$:$-\frac{1}{2} y^{-3/2} \frac{dy}{dx} - \frac{1}{2} x^{-3/2} = 0$$-\frac{1}{2 y^{3/2}} \frac{dy}{dx} = \frac{1}{2 x^{3/2}}$$\frac{dy}{dx} = -\frac{2 y^{3/2}}{2 x^{3/2}}$$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{3/2}$

If $\sqrt{x} + \sqrt{y} = \sqrt{xy}$,then $\frac{dy}{dx} = $

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