The problem asks you to use the definition of the derivative to show that if $f(x) = \frac{1}{x}$, then $f'(x) = -\frac{1}{x^2}$.

To solve this, let's proceed with the formal definition of a derivative, which is:

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

Given $f(x) = \frac{1}{x}$, we substitute this into the definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

Now, let's simplify the expression inside the limit:

1. Get a common denominator for the numerator:

$\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x+h)}$

2. Substitute this back into the limit expression:

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

3. Simplify the fraction:

$f'(x) = \lim_{h \to 0} \frac{-h}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-1}{x(x+h)}$

4. Take the limit as $h \to 0$:

$f'(x) = \frac{-1}{x(x+0)} = \frac{-1}{x^2}$

Thus, the derivative of $f(x) = \frac{1}{x}$ is indeed:

$f'(x) = -\frac{1}{x^2}$

Would you like further details or have any questions about this process?

Here are 5 follow-up questions you could explore:

1. How does the definition of the derivative relate to the concept of the slope of a tangent line?
2. What other methods, apart from the limit definition, can be used to compute derivatives?
3. How would the derivative change if $f(x) = \frac{1}{x^n}$ for a general $n$?
4. Can you explain how the result $f'(x) = -\frac{1}{x^2}$ behaves for different values of $x$?
5. How does this derivative relate to the power rule for negative exponents?

Tip: When dealing with fractions and derivatives, always remember that simplifying the algebra before taking the limit can save you time!