Implicit differentiation and evaluation at a point

Differentiate both sides with respect to $x$: $\frac{d}{dx}[x^{1/2}] + \frac{d}{dx}[y^{1/2}] = \frac{d}{dx}[4]$. Using the power rule and chain rule, this gives $\frac{1}{2}x^{-1/2} + \frac{1}{2}y^{-1/2}\cdot \frac{dy}{dx} = 0$.

Isolate $\frac{dy}{dx}$: $\frac{1}{2}y^{-1/2}\frac{dy}{dx} = -\frac{1}{2}x^{-1/2}$, so $\frac{dy}{dx} = -\frac{y^{-1/2}}{x^{-1/2}} = -\sqrt{\dfrac{y}{x}}$. Substituting $x=4$, $y=4$ gives $\frac{dy}{dx} = -\sqrt{4/4} = -1$.

The slope is negative. In part a, the derivative simplified to negative the square root of y over x, and evaluating at the point four four gave negative one. Since both x and y are positive on this curve, the square root term is always positive, so the negative sign out front makes the slope negative everywhere on the curve, including at four four.