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If $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{7 y-7 x}{y-7 x}$
(B)
Given : $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$
$\therefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}$
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. $x$
$\begin{array}{l}
2 x+2 y \frac{d y}{d x}=14\left[x \frac{d y}{d x}+y\right] \Rightarrow x+y \frac{d y}{d x}=7 x \cdot \frac{d y}{d x}+7 y \\
(y-7 x) \frac{d y}{d x}=7 y-x \Rightarrow \frac{d y}{d x}=\frac{7 y-x}{y-7 x}
\end{array}$
Given : $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$
$\therefore \frac{x+y}{\sqrt{x y}}=4 \Rightarrow x+y=4 \sqrt{x y}$
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. $x$
$\begin{array}{l}
2 x+2 y \frac{d y}{d x}=14\left[x \frac{d y}{d x}+y\right] \Rightarrow x+y \frac{d y}{d x}=7 x \cdot \frac{d y}{d x}+7 y \\
(y-7 x) \frac{d y}{d x}=7 y-x \Rightarrow \frac{d y}{d x}=\frac{7 y-x}{y-7 x}
\end{array}$
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