1. **State the problem:** Find the derivative of the function $$f(x) = x \sqrt{4 - x}$$. 2. **Recall the formula:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $$u(x) = x$$ and $$v(x) = \sqrt{4 - x} = (4 - x)^{1/2}$$. 3. **Differentiate each part:** - $$u'(x) = \frac{d}{dx}[x] = 1$$ - For $$v(x) = (4 - x)^{1/2}$$, use the chain rule: $$v'(x) = \frac{1}{2}(4 - x)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{4 - x}}$$ 4. **Apply the product rule:** $$f'(x) = u'(x)v(x) + u(x)v'(x) = 1 \cdot \sqrt{4 - x} + x \cdot \left(-\frac{1}{2\sqrt{4 - x}}\right)$$ 5. **Simplify the expression:** $$f'(x) = \sqrt{4 - x} - \frac{x}{2\sqrt{4 - x}}$$ 6. **Combine into a single fraction:** $$f'(x) = \frac{2(4 - x)}{2\sqrt{4 - x}} - \frac{x}{2\sqrt{4 - x}} = \frac{2(4 - x) - x}{2\sqrt{4 - x}}$$ 7. **Simplify numerator:** $$2(4 - x) - x = 8 - 2x - x = 8 - 3x$$ 8. **Final derivative:** $$f'(x) = \frac{8 - 3x}{2\sqrt{4 - x}}$$ This is the derivative of the given function.