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

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Aicharya · Accepted Answer

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, $(x+y)^{2}=16 x y \Rightarrow x^{2}+y^{2}=14 x y$ Differentiating both sides w.r.t. $x$ $\begin{array}{l} 2 x+2 y \frac{d y}{d x}=14\left[x \frac{d y}{d x}+yight] \Rightarrow x+y \frac{d y}{d x}=7 x \cdot \frac{d y}{d x}+7 y \ (y-7 x) \frac{d y}{d x}=7 y-x \Rightarrow \frac{d y}{d x}=\frac{7 y-x}{y-7 x} \end{array}$

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

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