1. **State the problem:** We want to find the derivative of the function $f(x) = \sqrt{4 - x}$ using the definition of the derivative. 2. **Recall the definition of the derivative:** The derivative of a function $f(x)$ at a point $x$ is given by $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This means we find the limit of the difference quotient as $h$ approaches zero. 3. **Apply the definition to $f(x) = \sqrt{4 - x}$:** $$f'(x) = \lim_{h \to 0} \frac{\sqrt{4 - (x+h)} - \sqrt{4 - x}}{h} = \lim_{h \to 0} \frac{\sqrt{4 - x - h} - \sqrt{4 - x}}{h}$$ 4. **Rationalize the numerator to simplify:** Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{4 - x - h} - \sqrt{4 - x}}{h} \cdot \frac{\sqrt{4 - x - h} + \sqrt{4 - x}}{\sqrt{4 - x - h} + \sqrt{4 - x}} = \frac{(4 - x - h) - (4 - x)}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)}$$ 5. **Simplify the numerator:** $$(4 - x - h) - (4 - x) = -h$$ So the expression becomes $$\frac{-h}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)}$$ 6. **Cancel $h$ in numerator and denominator:** $$\frac{-h}{h \left(\sqrt{4 - x - h} + \sqrt{4 - x}\right)} = \frac{-1}{\sqrt{4 - x - h} + \sqrt{4 - x}}$$ 7. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{4 - x - h} + \sqrt{4 - x}} = \frac{-1}{\sqrt{4 - x} + \sqrt{4 - x}} = \frac{-1}{2 \sqrt{4 - x}}$$ 8. **Final answer:** $$\boxed{f'(x) = -\frac{1}{2 \sqrt{4 - x}}}$$ This derivative tells us the rate of change of the function $\sqrt{4 - x}$ at any point $x$ where $4 - x > 0$ (i.e., $x < 4$).