We are given the function:

$f(x) = -\frac{1}{2x^3}$

We need to find the derivative of this function, $f'(x)$.

Step-by-step process:

1. Rewrite the function in a simpler form for differentiation:

   Instead of dealing with the fraction directly, rewrite $f(x)$ using exponents: $f(x) = -\frac{1}{2} x^{-3}$

2. Differentiate using the power rule:

   The power rule states that if $f(x) = ax^n$, then $f'(x) = a \cdot n \cdot x^{n-1}$.

   Here, $a = -\frac{1}{2}$ and $n = -3$.

   Apply the power rule: $f'(x) = -\frac{1}{2} \cdot (-3) \cdot x^{-3 - 1}$ $f'(x) = \frac{3}{2} x^{-4}$

3. Rewrite the derivative without negative exponents:

   Since $x^{-4} = \frac{1}{x^4}$, the derivative becomes: $f'(x) = \frac{3}{2x^4}$

Thus, the derivative of the function is:

$f'(x) = \frac{3}{2x^4}$

Would you like more details, or do you have any questions?

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Here are 5 related questions to expand your understanding:

1. How do we apply the product rule if two functions are multiplied?
2. How would the derivative change if the exponent of $x$ were positive instead of negative?
3. What happens to the derivative if we multiply the function by a constant?
4. How do we handle the chain rule in functions with more complex exponents?
5. What is the second derivative of $f(x) = -\frac{1}{2x^3}$?

Tip: Always simplify the expression before differentiating to avoid confusion with fractions or complex powers.