1. **State the problem:** Find the derivative of the function $y = 8 \sin \sqrt{u}$ with respect to $u$. 2. **Recall the formula:** To differentiate $y = 8 \sin \sqrt{u}$, we use the chain rule. The chain rule states that if $y = f(g(u))$, then $$\frac{dy}{du} = f'(g(u)) \cdot g'(u).$$ 3. **Identify inner and outer functions:** Here, the outer function is $f(x) = 8 \sin x$ and the inner function is $g(u) = \sqrt{u} = u^{1/2}$. 4. **Differentiate the outer function:** The derivative of $8 \sin x$ with respect to $x$ is $$8 \cos x.$$ So, $$f'(g(u)) = 8 \cos \sqrt{u}.$$ 5. **Differentiate the inner function:** The derivative of $g(u) = u^{1/2}$ is $$g'(u) = \frac{1}{2} u^{-1/2} = \frac{1}{2 \sqrt{u}}.$$ 6. **Apply the chain rule:** Multiply the derivatives: $$\frac{dy}{du} = 8 \cos \sqrt{u} \times \frac{1}{2 \sqrt{u}} = \frac{8}{2 \sqrt{u}} \cos \sqrt{u} = \frac{4 \cos \sqrt{u}}{\sqrt{u}}.$$ 7. **Final answer:** $$\boxed{\frac{dy}{du} = \frac{4 \cos \sqrt{u}}{\sqrt{u}}}.$$