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

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(B) Given that $\sqrt{x+y}-\sqrt{x-y}=c$. Squaring both sides, we get $(x+y)+(x-y)-2 \sqrt{x^2-y^2}=c^2$. $2x - c^2 = 2 \sqrt{x^2-y^2}$. Squaring both

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$\frac{-c^4}{4 y^3}$

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(B) Given that $\sqrt{x+y}-\sqrt{x-y}=c$. Squaring both sides, we get $(x+y)+(x-y)-2 \sqrt{x^2-y^2}=c^2$. $2x - c^2 = 2 \sqrt{x^2-y^2}$. Squaring both sides again, we get $4x^2 + c^4 - 4xc^2 = 4(x^2-y^2) = 4x^2 - 4y^2$. $c^4 - 4xc^2 = -4y^2$, which implies $4y^2 = 4xc^2 - c^4$. Differentiating with respect to $x$, we get $8y \frac{dy}{dx} = 4c^2$, so $2y \frac{dy}{dx} = c^2$. Differentiating again with respect to $x$, we get $2(\frac{dy}{dx})^2 + 2y \frac{d^2y}{dx^2} = 0$. Thus, $\frac{d^2y}{dx^2} = -\frac{(\frac{dy}{dx})^2}{y}$. Since $\frac{dy}{dx} = \frac{c^2}{2y}$, we have $\frac{d^2y}{dx^2} = -\frac{(c^2/2y)^2}{y} = -\frac{c^4}{4y^3}$. Therefore, option $B$ is correct.

If $\sqrt{x+y}-\sqrt{x-y}=c$, then $\frac{d^2 y}{d x^2}=$

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