To find the derivative of the function $f(x) = x + \frac{1}{x}$, we can apply basic differentiation rules.

1. Differentiate each term separately:

   $\frac{d}{dx}\left(x + \frac{1}{x}\right) = \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{1}{x}\right)$

2. Apply the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ to each term:

   • For the first term, $\frac{d}{dx}(x) = 1$.
   • For the second term, rewrite $\frac{1}{x}$ as $x^{-1}$ and differentiate: $\frac{d}{dx}\left(x^{-1}\right) = -x^{-2} = -\frac{1}{x^2}$

3. Combine the results:

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

Final Answer

The derivative of $f(x) = x + \frac{1}{x}$ is: $f'(x) = 1 - \frac{1}{x^2}$

Would you like a more detailed explanation on any step, or have any questions?

Here are some related questions to further your understanding:

1. How do you find the second derivative of $f(x) = x + \frac{1}{x}$?
2. What are critical points, and how would you find them for $f(x) = x + \frac{1}{x}$?
3. How can you determine if $f(x) = x + \frac{1}{x}$ has any inflection points?
4. What is the derivative of $g(x) = x - \frac{1}{x}$?
5. How can you apply this derivative to analyze the behavior of $f(x)$ at certain values of $x$?

Tip: Remember, rewriting terms with negative exponents (e.g., $\frac{1}{x} = x^{-1}$) can often make differentiation easier.