To find the derivative of $f(x) = \frac{1}{x}$ using first principles (i.e., the definition of the derivative), we use the formula:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 1: Write $f(x)$ and $f(x+h)$

Given that $f(x) = \frac{1}{x}$, we calculate $f(x+h)$: $f(x+h) = \frac{1}{x+h}$

Step 2: Apply the formula

Substitute $f(x) = \frac{1}{x}$ and $f(x+h) = \frac{1}{x+h}$ into the definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

Step 3: Simplify the expression

Combine the two fractions in the numerator:

= \lim_{h \to 0} \frac{\frac{-h}{(x+h)x}}{h}$$ ### Step 4: Simplify further $$f'(x) = \lim_{h \to 0} \frac{-h}{h(x+h)x}$$ Cancel the $$h$$'s in the numerator and denominator: $$f'(x) = \lim_{h \to 0} \frac{-1}{(x+h)x}$$ ### Step 5: Take the limit as $$h \to 0$$ As $$h$$ approaches 0, $$x+h$$ approaches $$x$$, so the expression simplifies to: $$f'(x) = \frac{-1}{x^2}$$ Thus, the derivative of $$f(x) = \frac{1}{x}$$ is: $$f'(x) = \frac{-1}{x^2}$$ Would you like any further details or clarifications? --- Here are 5 related questions for further practice: 1. Use first principles to differentiate $$f(x) = x^2$$. 2. Find the derivative of $$f(x) = \frac{1}{x^2}$$ using first principles. 3. What is the second derivative of $$f(x) = \frac{1}{x}$$? 4. Differentiate $$f(x) = \ln(x)$$ using the definition of a derivative. 5. Use first principles to find the derivative of $$f(x) = \frac{1}{x+1}$$. **Tip:** Mastering first principles helps in understanding how all differentiation rules are derived, enhancing your problem-solving flexibility.