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

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(D) Given equation is $\sqrt{\frac{y}{x}}+4 \sqrt{\frac{x}{y}}=4$.Let $u = \sqrt{\frac{y}{x}}$. Then the equation becomes $u + \frac{4}{u} = 4$.Multip

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$4$

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(D) Given equation is $\sqrt{\frac{y}{x}}+4 \sqrt{\frac{x}{y}}=4$.Let $u = \sqrt{\frac{y}{x}}$. Then the equation becomes $u + \frac{4}{u} = 4$.Multiplying by $u$, we get $u^2 - 4u + 4 = 0$, which is $(u-2)^2 = 0$.Thus, $u = 2$, which implies $\sqrt{\frac{y}{x}} = 2$.Squaring both sides, we get $\frac{y}{x} = 4$, or $y = 4x$.Now, differentiating both sides with respect to $x$, we get $\frac{d y}{d x} = \frac{d}{d x}(4x) = 4$.Therefore, $\frac{d y}{d x} = 4$.

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

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