1. **State the problem:** We need to find the derivative of the function $$f(x) = \frac{x}{x+1}$$ using the definition of the derivative. 2. **Recall the definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition:** $$f'(x) = \lim_{h \to 0} \frac{\frac{x+h}{(x+h)+1} - \frac{x}{x+1}}{h} = \lim_{h \to 0} \frac{\frac{x+h}{x+h+1} - \frac{x}{x+1}}{h}$$ 4. **Find a common denominator for the 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 the numerator:** $$(x+h)(x+1) = x^2 + x + hx + h$$ $$x(x+h+1) = x^2 + xh + x$$ 6. **Subtract the two expressions:** $$x^2 + x + hx + h - (x^2 + xh + x) = x^2 + x + hx + h - x^2 - xh - x = h$$ 7. **So the numerator simplifies to:** $$\frac{h}{(x+h+1)(x+1)}$$ 8. **Substitute back into the difference quotient:** $$f'(x) = \lim_{h \to 0} \frac{\frac{h}{(x+h+1)(x+1)}}{h} = \lim_{h \to 0} \frac{h}{(x+h+1)(x+1)} \cdot \frac{1}{h}$$ 9. **Cancel the $h$ terms:** $$= \lim_{h \to 0} \frac{\cancel{h}}{(x+h+1)(x+1)} \cdot \frac{1}{\cancel{h}} = \lim_{h \to 0} \frac{1}{(x+h+1)(x+1)}$$ 10. **Evaluate the limit as $h \to 0$:** $$f'(x) = \frac{1}{(x+1)(x+1)} = \frac{1}{(x+1)^2}$$ **Final answer:** $$\boxed{f'(x) = \frac{1}{(x+1)^2}}$$