1. **State the problem:** Find the derivative $\frac{dy}{dx}$ for the equation $$\sqrt{x} + \sqrt{y} = 4.$$\n\n2. **Recall the formula and rules:** We will use implicit differentiation since $y$ is a function of $x$. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, and for $\sqrt{y}$, by chain rule, it is $\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx}$.\n\n3. **Differentiate both sides with respect to $x$:$$\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(4)$$which gives$$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0.$$\n\n4. **Solve for $\frac{dy}{dx}$:**\n$$\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$$\nMultiply both sides by $2\sqrt{y}$:\n$$\frac{dy}{dx} = -\frac{2\sqrt{y}}{2\sqrt{x}} = -\frac{\sqrt{y}}{\sqrt{x}}.$$\n\n5. **Final answer:**\n$$\boxed{\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}}.$$\nThis means the rate of change of $y$ with respect to $x$ depends on the ratio of the square roots of $y$ and $x$, with a negative sign indicating opposite directions of change.