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

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Aicharya · Accepted Answer

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. $x$, $10 	imes \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}ight)-2 	imes 1=0 \Rightarrow \frac{5}{\sqrt{x+y}}\left(1+\frac{d y}{d x}ight)=2$ $	herefore 1+\frac{d y}{d x}=\frac{2 \sqrt{x+y}}{5}....(1)$ Differentiating w.r.t. $x$, $\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{2}{5} 	imes \frac{1}{2 \sqrt{x+y}}\left(1+\frac{d y}{d x}ight)=\frac{1}{5 \sqrt{x+y}} 	imes \frac{2 \sqrt{x+y}}{5} \quad \ldots[\because 	ext { From (1) }] \ &=\frac{2}{25} \end{aligned}$

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

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