1. The problem is to find the derivative of the function $$f(x) = x \sqrt{4 - x}$$ and explain how to add fractions together into one fraction during the process. 2. First, recall the product rule for derivatives: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = x$ and $v(x) = \sqrt{4 - x}$. 3. Compute the derivatives: - $$u'(x) = 1$$ - To find $$v'(x)$$, rewrite $$v(x) = (4 - x)^{1/2}$$ and 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: $$f'(x) = 1 \cdot \sqrt{4 - x} + x \cdot \left(-\frac{1}{2\sqrt{4 - x}}\right) = \sqrt{4 - x} - \frac{x}{2\sqrt{4 - x}}$$ 5. To add these two terms into one fraction, find a common denominator. The first term $$\sqrt{4 - x}$$ can be written as $$\frac{\sqrt{4 - x} \cdot 2\sqrt{4 - x}}{2\sqrt{4 - x}} = \frac{2(4 - x)}{2\sqrt{4 - x}}$$. 6. Now add the fractions: $$\frac{2(4 - x)}{2\sqrt{4 - x}} - \frac{x}{2\sqrt{4 - x}} = \frac{2(4 - x) - x}{2\sqrt{4 - x}}$$ 7. Simplify the numerator: $$2(4 - x) - x = 8 - 2x - x = 8 - 3x$$ 8. The derivative in one fraction is: $$f'(x) = \frac{8 - 3x}{2\sqrt{4 - x}}$$ This shows how to add fractions by finding a common denominator and combining the numerators.