Question

If $y\sqrt{1+x}+x\sqrt{1+y}=0$ then value of $\frac{dy}{dx}$ at $y=1$ is,

• A. $-\frac{1}{2}$
• B. $1$
• C. $-4$
• D. $2$

Answer 1

The correct option is C $-4$

$y\sqrt{1+x}+x\sqrt{1+y}=0$ Given
Differentiating w.r.t $x$
$\Rightarrow \left(\frac{dy}{dx}\right)\sqrt{1+x}+\frac{1}{2\sqrt{1+x}}y+\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}\frac{dy}{dx}=0$
at $y=1,x=-\frac{1}{2}$

$\therefore \left(\frac{dy}{dx}\right)\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}+\sqrt{2}-\frac{1}{4\sqrt{2}}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-4$