Given the function f, of, x, equals, minus, start fraction, 1, divided by, 2, x, cubed, end fraction, commaf(x)=− 
2x 
3
 
1
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 , find f, prime, of, x, .f 
′
 (x). Express your answer in simplest form without using negative exponents.

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.

Given the function f(x) = -1/(2x^3), find f'(x). Express your answer in simplest form without using negative exponents.

Find the derivative of f(x) = -1/(2x^3) using the power rule. This step-by-step solution simplifies the expression and applies calculus techniques. Suitable for students in Grades 11-12 or early college courses.

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