If x+y+y-x=c, then d2ydx2 equals

We have, x+y+y-x=cDifferentiating with respect to x, we obtain 1x+y1+dydx+1y−xdydx−1=0⇒ dydx(y−x+y+x)=y+x−y−x⇒ edydx=y+x−y−x⇒ e2dydx2=(y+x−y−x)2 …(i)But, (y+x+y−x)2+(y+x−y−x)2=4y⇒ (y+x−y−x)2=4y−c2 …(ii)From (i) and (ii), we obtain e2dydx2=4y−c2Differentiating with respect to x, we obtain 2e2dydxd2ydx2=4⋅dydx⇒ d2ydx2=2e2

If x+y+y-x=c, then d2ydx2 equals

We have, x+y+y-x=cDifferentiating with respect to x, we obtain 1x+y1+dydx+1y−xdydx−1=0⇒ dydx(y−x+y+x)=y+x−y−x⇒ edydx=y+x−y−x⇒ e2dydx2=(y+x−y−x)2 …(i)But, (y+x+y−x)2+(y+x−y−x)2=4y⇒ (y+x−y−x)2=4y−c2 …(ii)From (i) and (ii), we obtain e2dydx2=4y−c2Differentiating with respect to x, we obtain 2e2dydxd2ydx2=4⋅dydx⇒ d2ydx2=2e2

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

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$2 / c^{2}$

$2 / c$

$- 2 / c^{2}$

$- 2 / c$

answer is C.

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Detailed Solution

We have,

$\sqrt{x + y} + \sqrt{y - x} = c$

Differentiating with respect to $x,$ we obtain

$\begin{array}{l} \frac{1}{\sqrt{x} + y} (1 + \frac{d y}{d x}) + \frac{1}{\sqrt{y - x}} (\frac{d y}{d x} - 1) = 0 \\ \Rightarrow \frac{d y}{d x} (\sqrt{y - x} + \sqrt{y + x}) = \sqrt{y + x} - \sqrt{y - x} \\ \Rightarrow e \frac{d y}{d x} = \sqrt{y + x} - \sqrt{y - x} \end{array}$

$\Rightarrow e^{2} {(\frac{d y}{d x})}^{2} = (\sqrt{y + x} - \sqrt{y - x})^{2}$ …(i)

But, $(\sqrt{y + x} + \sqrt{y - x})^{2} + (\sqrt{y + x} - \sqrt{y - x})^{2} = 4 y$

$\Rightarrow (\sqrt{y + x} - \sqrt{y - x})^{2} = 4 y - c^{2}$ …(ii)

From (i) and (ii), we obtain

$e^{2} {(\frac{d y}{d x})}^{2} = 4 y - c^{2}$

Differentiating with respect to $x,$ we obtain

$\begin{array}{l} 2 e^{2} \frac{d y}{d x} \frac{d^{2} y}{d x^{2}} = 4 \cdot \frac{d y}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{2}{e^{2}} \end{array}$