Question

$\sqrt{x} + \sqrt{y} \Rightarrow \sqrt{xy}$ $\frac{dy}{dx} \Rightarrow ?$

Answer 1

$\frac{dy}{dx} = -\frac{1}{\left(\sqrt{x}-1\right)^3}$

Answer 2

1. Rewrite $\sqrt{x}+\sqrt{y}=\sqrt{xy}$ using $u=\sqrt{x}$ and $v=\sqrt{y}$ to get $(u-1)(v-1)=1$.

2. Differentiate $\sqrt{x}+\sqrt{y}=\sqrt{xy}$ implicitly to obtain:

   $$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=\frac{1}{2\sqrt{xy}}\left(y+x\frac{dy}{dx}\right).$$

3. Simplify and collect $\frac{dy}{dx}$ terms to get a relation.

4. Using the relation from the factorization, find the explicit derivative:

   $$\frac{dy}{dx} = -\frac{1}{(\sqrt{x}-1)^3}.$$