1. **State the problem:** Differentiate the function $y = (4x)^{0.5}$ with respect to $x$. 2. **Recall the formula:** The derivative of $x^n$ with respect to $x$ is given by $$\frac{d}{dx} x^n = n x^{n-1}.$$ Also, the chain rule states that $$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x).$$ 3. **Apply the chain rule:** Here, $y = (4x)^{0.5}$ can be seen as $y = u^{0.5}$ where $u = 4x$. 4. **Differentiate the outer function:** $$\frac{dy}{du} = 0.5 u^{-0.5} = \frac{0.5}{\sqrt{u}}.$$ 5. **Differentiate the inner function:** $$\frac{du}{dx} = 4.$$ 6. **Combine using the chain rule:** $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{0.5}{\sqrt{4x}} \cdot 4.$$ 7. **Simplify:** $$\frac{dy}{dx} = \frac{0.5 \times 4}{\sqrt{4x}} = \frac{2}{\sqrt{4x}}.$$ 8. **Simplify the denominator:** $$\sqrt{4x} = \sqrt{4} \times \sqrt{x} = 2 \sqrt{x}.$$ 9. **Final simplification:** $$\frac{dy}{dx} = \frac{2}{2 \sqrt{x}} = \frac{\cancel{2}}{\cancel{2} \sqrt{x}} = \frac{1}{\sqrt{x}}.$$ **Answer:** $$\frac{dy}{dx} = \frac{1}{\sqrt{x}}.$$