Question

If $\sqrt{x}+\sqrt{y}=4$ then find $\frac{dy}{dx}$ at $x=1$.

Answer 1

Given,

$\sqrt{x}+\sqrt{y}=4$.......(1).

Now differentiating both sides with respect to $x$ we get,

$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

or, $\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$

or, ${\frac{dy}{dx}|}_{x=1}=-\frac{3}{1}=-3$ [ When $x=1$ then form (1) we get $\sqrt{y}=3$.]