Question:

Let $y = \sqrt{x^2 + 16}$ and $s = \frac{x}{y}$. Differentiating $y$ with respect to $x$: $\frac{dy}{dx} = \frac{x}{\sqrt{x^2+16}}$. For $s = \frac{x}{y}$, we have $\frac{ds}{dx} = \frac{y - x\frac{dy}{dx}}{y^2} = \frac{1-x^2}{(x^2+16)^{3/2}}$. At $x = 3$: $\frac{ds}{dx} = \frac{1-9}{(9+16)^{3/2}} = \frac{-8}{125} = -\frac{12}{5}$.