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