Question

If $x\sqrt{1+y}+y\sqrt{1+x}=0;x\ne y$, then the value of $\frac{d^2y}{dx^2}+2\frac{dy}{dx}$ at x = 1 is :

• A. 0
• B. $\frac{1}{4}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$-\frac{1}{4}$

Answer 2

The correct option is (C) :$-\frac{1}{4}$.