If sqrt{frac{x}{y}} + sqrt{frac{y}{x}} = 6, t | vedclass

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Question

(B) Given: $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$. Multiply by $\sqrt{xy}$: $x + y = 6\sqrt{xy}$. Square both sides: $(x + y)^2 = 36xy$. $x^2 +

Vedclass · Accepted Answer

$\frac{x - 17y}{17x - y}$

Vedclass · Answer

(B) Given: $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$. Multiply by $\sqrt{xy}$: $x + y = 6\sqrt{xy}$. Square both sides: $(x + y)^2 = 36xy$. $x^2 + 2xy + y^2 = 36xy \implies x^2 - 34xy + y^2 = 0$. Differentiate with respect to $x$: $\frac{d}{dx}(x^2) - 34\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$. $2x - 34(y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$. Divide by $2$: $x - 17y - 17x\frac{dy}{dx} + y\frac{dy}{dx} = 0$. $x - 17y = \frac{dy}{dx}(17x - y)$. $\frac{dy}{dx} = \frac{x - 17y}{17x - y}$.

If $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 6$, then $\frac{dy}{dx} = $

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