1. **State the problem:** Given the function $y = \frac{8}{\sqrt{x}}$, find the derivative $y'(4)$.\n\n2. **Rewrite the function:** Recall that $\sqrt{x} = x^{1/2}$, so we can write \n$$y = 8x^{-1/2}.$$\n\n3. **Differentiate the function:** Using the power rule $\frac{d}{dx} x^n = nx^{n-1}$, we get \n$$y' = 8 \times \left(-\frac{1}{2}\right) x^{-3/2} = -4 x^{-3/2}.$$\n\n4. **Evaluate the derivative at $x=4$:** \n$$y'(4) = -4 \times 4^{-3/2}.$$\nCalculate $4^{-3/2}$: \n$$4^{1/2} = 2 \implies 4^{-3/2} = \frac{1}{4^{3/2}} = \frac{1}{(4^{1/2})^3} = \frac{1}{2^3} = \frac{1}{8}.$$\nSo, \n$$y'(4) = -4 \times \frac{1}{8} = -\frac{1}{2}.$$\n\n**Final answer:** $y'(4) = -\frac{1}{2}$.