1. **State the problem:** Find the derivative of the function $f(x) = 5x - x^2$ using the difference quotient method. 2. **Recall the difference quotient formula for the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Calculate $f(x+h)$:** $$f(x+h) = 5(x+h) - (x+h)^2 = 5x + 5h - (x^2 + 2xh + h^2) = 5x + 5h - x^2 - 2xh - h^2$$ 4. **Form the difference quotient:** $$\frac{f(x+h) - f(x)}{h} = \frac{(5x + 5h - x^2 - 2xh - h^2) - (5x - x^2)}{h}$$ 5. **Simplify the numerator:** $$5x + 5h - x^2 - 2xh - h^2 - 5x + x^2 = 5h - 2xh - h^2$$ 6. **Rewrite the difference quotient:** $$\frac{5h - 2xh - h^2}{h} = \frac{h(5 - 2x - h)}{h}$$ 7. **Cancel $h$ (for $h \neq 0$):** $$5 - 2x - h$$ 8. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} (5 - 2x - h) = 5 - 2x$$ **Final answer:** $$f'(x) = 5 - 2x$$ This means the slope of the tangent line to the curve $f(x) = 5x - x^2$ at any point $x$ is $5 - 2x$.