If sqrt{frac{x}{y}}+sqrt{frac{y}{x}}=4,then f | vedclass

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(B) Given: $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$ $	herefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}$ Squaring both sides,we get: $(

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$\frac{7 y-x}{y-7 x}$

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(B) Given: $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$ $	herefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}$ Squaring both sides,we get: $(x+y)^{2}=16 x y \Rightarrow x^{2}+2 x y+y^{2}=16 x y \Rightarrow x^{2}+y^{2}=14 x y$ Differentiating both sides with respect to $x$: $2 x+2 y \frac{d y}{d x}=14 \left(x \frac{d y}{d x}+yight)$ Dividing by $2$: $x+y \frac{d y}{d x}=7 x \frac{d y}{d x}+7 y$ Rearranging the terms to solve for $\frac{d y}{d x}$: $y \frac{d y}{d x}-7 x \frac{d y}{d x}=7 y-x$ $(y-7 x) \frac{d y}{d x}=7 y-x$ $\frac{d y}{d x}=\frac{7 y-x}{y-7 x}$

If $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$,then $\frac{d y}{d x}=$

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