If sqrt{x+y}+sqrt{y-x}=5,then left(frac{d^{2} | vedclass

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(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$. Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$. Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x

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$\frac{2}{25}$

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(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$. Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$. Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x+y}$. Simplifying,$-2x = 25 - 10\sqrt{x+y}$,which gives $10\sqrt{x+y} = 2x + 25$. Differentiating both sides with respect to $x$: $10 	imes \frac{1}{2\sqrt{x+y}} 	imes (1 + \frac{dy}{dx}) = 2$. $\frac{5}{\sqrt{x+y}} 	imes (1 + \frac{dy}{dx}) = 2$. $1 + \frac{dy}{dx} = \frac{2\sqrt{x+y}}{5}$. Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{2}{5} 	imes \frac{1}{2\sqrt{x+y}} 	imes (1 + \frac{dy}{dx})$. Substituting the value of $(1 + \frac{dy}{dx})$ from the previous step: $\frac{d^2y}{dx^2} = \frac{1}{5\sqrt{x+y}} 	imes \frac{2\sqrt{x+y}}{5} = \frac{2}{25}$.

If $\sqrt{x+y}+\sqrt{y-x}=5$,then $\left(\frac{d^{2} y}{d x^{2}}\right)=$

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