We can solve the derivative of the function $\frac{s}{(s^2 - a^2)^3}$ using the quotient rule, since it's a ratio of two functions. The quotient rule is:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

For $f(s) = s$ and $g(s) = (s^2 - a^2)^3$, we can proceed as follows:

Step 1: Derivatives of $f(s)$ and $g(s)$

• $f(s) = s$, so $f'(s) = 1$
• $g(s) = (s^2 - a^2)^3$, so we need to apply the chain rule to find $g'(s)$.

To find $g'(s)$, use the chain rule: $g'(s) = 3(s^2 - a^2)^2 \cdot 2s = 6s(s^2 - a^2)^2$

Step 2: Apply the quotient rule

Now apply the quotient rule:

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

Step 3: Simplify the expression

Simplify the numerator: $= \frac{(s^2 - a^2)^3 - 6s^2(s^2 - a^2)^2}{(s^2 - a^2)^6}$

Factor out $(s^2 - a^2)^2$ from the numerator: $= \frac{(s^2 - a^2)^2 \left( (s^2 - a^2) - 6s^2 \right)}{(s^2 - a^2)^6}$

Simplify further: $= \frac{(s^2 - a^2)^2 \left( -5s^2 - a^2 \right)}{(s^2 - a^2)^6}$

Cancel out $(s^2 - a^2)^2$ from the numerator and denominator: $= \frac{-5s^2 - a^2}{(s^2 - a^2)^4}$

Final Answer:

$\frac{d}{ds} \left( \frac{s}{(s^2 - a^2)^3} \right) = \frac{-5s^2 - a^2}{(s^2 - a^2)^4}$

Let me know if you'd like more details on any of these steps!

5 Related Questions:

1. What is the chain rule in calculus, and how is it applied in derivative problems?
2. Can you explain how the quotient rule works with different examples?
3. How do you find higher-order derivatives of more complex rational functions?
4. What are some common mistakes to avoid when using the quotient rule?
5. How can the derivative of implicit functions differ from explicit ones?

Tip:

Always simplify the expression as much as possible before substituting into the quotient rule—it makes the differentiation process smoother!