Question

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

• A. $\frac{2}{c^2}$
• B. $-\frac{2}{c^2}$
• C. $\frac{2}{c}$
• D. $-\frac{2}{c}$

Answer 1

The correct option is A $\frac{2}{c^2}$

Given: $\sqrt{x+y}+\sqrt{y-x}=c$ ...(1)

$\therefore \frac{1}{\sqrt{x+y}+\sqrt{y-x}}=\frac{1}{c}$

Rationalising the denominator, we get

$\Rightarrow \frac{\sqrt{x+y}-\sqrt{y-x}}{(x+y)-(y-x)}=\frac{1}{c}$

$\Rightarrow \frac{\sqrt{x+y}-\sqrt{y-x}}{2x}=\frac{1}{c}$

$\Rightarrow \sqrt{x+y}-\sqrt{y-x}=\frac{2x}{c}$ ...(2)

By adding (1) and (2) we get

$2\sqrt{y+x}=c+\frac{2x}{c}$

$\Rightarrow 4(y+x)=c^2+\frac{{4x}^2}{c^2}+4x$

$\therefore 4(\frac{dy}{dx}+1)=\frac{8x}{c^2}+4$

$\therefore 4\frac{dy}{dx}=\frac{8x}{c^2}$

$\therefore \frac{d^{2}y}{{dx}^2}=\frac{2}{c^2}$