We are asked to differentiate the function:

\[
f(s) = \frac{s}{(s^2 + a^2)^3}
\]

To find \(\frac{d}{ds}\left( \frac{s}{(s^2 + a^2)^3} \right)\), we will apply the **quotient rule**. The quotient rule for differentiation states:

\[
\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}
\]

### Step 1: Identify \(u(s)\) and \(v(s)\)
Here, 
- \( u(s) = s \) 
- \( v(s) = (s^2 + a^2)^3 \)

### Step 2: Differentiate \(u(s)\) and \(v(s)\)
- \( u'(s) = \frac{d}{ds}(s) = 1 \)
- \( v'(s) = \frac{d}{ds} \left( (s^2 + a^2)^3 \right) \)

For \(v'(s)\), we apply the **chain rule**:
\[
v'(s) = 3(s^2 + a^2)^2 \cdot \frac{d}{ds}(s^2 + a^2) = 3(s^2 + a^2)^2 \cdot 2s = 6s(s^2 + a^2)^2
\]

### Step 3: Apply the quotient rule
Using the quotient rule:

\[
\frac{d}{ds}\left( \frac{s}{(s^2 + a^2)^3} \right) = \frac{(s^2 + a^2)^3 \cdot 1 - s \cdot 6s(s^2 + a^2)^2}{(s^2 + a^2)^6}
\]

### Step 4: Simplify the expression
Simplify the numerator:

\[
(s^2 + a^2)^3 - 6s^2(s^2 + a^2)^2
\]

Factor out \((s^2 + a^2)^2\) from both terms:

\[
(s^2 + a^2)^2 \left[ (s^2 + a^2) - 6s^2 \right] = (s^2 + a^2)^2 (a^2 - 5s^2)
\]

Thus, the derivative is:

\[
\frac{(s^2 + a^2)^2 (a^2 - 5s^2)}{(s^2 + a^2)^6} = \frac{a^2 - 5s^2}{(s^2 + a^2)^4}
\]

### Final Answer:
\[
\frac{d}{ds} \left( \frac{s}{(s^2 + a^2)^3} \right) = \frac{a^2 - 5s^2}{(s^2 + a^2)^4}
\]

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Would you like more details, or have any questions?

Here are some related questions:
1. What are the steps for using the product rule instead of quotient rule?
2. How can we apply the chain rule to functions with higher powers?
3. What happens when \(a = 0\) in this function?
4. How does the quotient rule differ from the chain rule?
5. Can you find the second derivative of the function?

**Tip:** In complex derivatives, always simplify each term before attempting further steps.

Learn how to differentiate the function s/(s^2 + a^2)^3 using the quotient and chain rules. This step-by-step guide walks through identifying u(s) and v(s), applying the quotient rule, and simplifying the derivative. Suitable for undergraduate calculus students.

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