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

(D)
Given \(\sqrt{x+y}+\sqrt{y-x}=5 \Rightarrow \sqrt{y-x}=5-\sqrt{x+y}\)
On squaring both side, we get
\(
y-x=25+x+y-10 \sqrt{x+y} \Rightarrow 10 \sqrt{x+y}-2 x=25
\)
Differentiating w.r.t. \(x\),
\(10 \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)-2 \times 1=0 \Rightarrow \frac{5}{\sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=2\)
\(
\therefore 1+\frac{d y}{d x}=\frac{2 \sqrt{x+y}}{5}....(1)
\)
Differentiating w.r.t. \(x\),
\(\frac{d^{2} y}{d x^{2}} =\frac{2}{5} \times \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}\right)=\frac{1}{5 \sqrt{x+y}}\) \(\times \frac{2 \sqrt{x+y}}{5} \quad \ldots[\because \text { From (1) }] \)
\( =\frac{2}{25}\)