1. **Problem Statement:** Differentiate the function $f(x) = \frac{x}{x-1}$ using the definition of the derivative. 2. **Definition of Derivative:** The derivative $f'(x)$ is defined as $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the function:** Substitute $f(x+h) = \frac{x+h}{(x+h)-1} = \frac{x+h}{x+h-1}$ and $f(x) = \frac{x}{x-1}$ into the limit: $$f'(x) = \lim_{h \to 0} \frac{\frac{x+h}{x+h-1} - \frac{x}{x-1}}{h}$$ 4. **Find common denominator and simplify numerator:** $$\frac{x+h}{x+h-1} - \frac{x}{x-1} = \frac{(x+h)(x-1) - x(x+h-1)}{(x+h-1)(x-1)}$$ 5. **Expand numerator:** $$(x+h)(x-1) = x^2 - x + hx - h$$ $$x(x+h-1) = x^2 + hx - x$$ 6. **Subtract the two expressions:** $$ (x^2 - x + hx - h) - (x^2 + hx - x) = x^2 - x + hx - h - x^2 - hx + x = (-x + x) + (hx - hx) + (-h) = -h$$ 7. **Rewrite the difference quotient:** $$\frac{\frac{x+h}{x+h-1} - \frac{x}{x-1}}{h} = \frac{\frac{-h}{(x+h-1)(x-1)}}{h} = \frac{-h}{h (x+h-1)(x-1)} = \frac{-1}{(x+h-1)(x-1)}$$ 8. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} \frac{-1}{(x+h-1)(x-1)} = \frac{-1}{(x-1)^2}$$ **Final answer:** $$f'(x) = \frac{-1}{(x-1)^2}$$ This means the slope of the function $f(x)$ at any point $x$ is $\frac{-1}{(x-1)^2}$. The negative sign indicates the function is decreasing where defined, and the denominator squared means the slope becomes very steep near $x=1$ where the function has a vertical asymptote.