1. **Problem Statement:** Find the derivative of the function $f(x) = x + \frac{1}{x}$ using the definition of the derivative. 2. **Definition of 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 formula calculates the slope of the tangent line to the curve at the point $x$. 3. **Apply the Definition:** $$f'(x) = \lim_{h \to 0} \frac{(x+h) + \frac{1}{x+h} - \left(x + \frac{1}{x}\right)}{h}$$ 4. **Simplify the Expression:** $$= \lim_{h \to 0} \frac{x + h + \frac{1}{x+h} - x - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{h + \frac{1}{x+h} - \frac{1}{x}}{h}$$ 5. **Combine the Fraction Terms:** $$= \lim_{h \to 0} \frac{h + \frac{x - (x+h)}{x(x+h)}}{h} = \lim_{h \to 0} \frac{h + \frac{-h}{x(x+h)}}{h}$$ 6. **Factor $h$ in the Numerator:** $$= \lim_{h \to 0} \frac{h \left(1 - \frac{1}{x(x+h)}\right)}{h} = \lim_{h \to 0} \left(1 - \frac{1}{x(x+h)}\right)$$ 7. **Evaluate the Limit as $h \to 0$:** $$= 1 - \frac{1}{x \cdot x} = 1 - \frac{1}{x^2}$$ 8. **Final Answer:** $$f'(x) = 1 - \frac{1}{x^2}$$ This derivative tells us the slope of the tangent line to the curve $f(x) = x + \frac{1}{x}$ at any point $x$.