To find the derivative of $\frac{\sqrt{3 - x^2}}{3 + x}$, we'll need to apply both the quotient rule and the chain rule.

Steps:

1. Let $f(x) = \sqrt{3 - x^2}$ and $g(x) = 3 + x$.
2. Use the quotient rule, which states:

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

3. Find $f'(x)$ using the chain rule for $f(x) = (3 - x^2)^{1/2}$:

$f'(x) = \frac{1}{2}(3 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{3 - x^2}}$

4. Find $g'(x)$:

$g'(x) = 1$

5. Apply the quotient rule:

$\left( \frac{\sqrt{3 - x^2}}{3 + x} \right)' = \frac{(3 + x) \cdot \frac{-x}{\sqrt{3 - x^2}} - \sqrt{3 - x^2} \cdot 1}{(3 + x)^2}$

6. Simplify the expression:

$= \frac{(3 + x) \cdot \frac{-x}{\sqrt{3 - x^2}} - \sqrt{3 - x^2}}{(3 + x)^2}$

7. Combine terms:

$= \frac{-x(3 + x) - (3 - x^2)}{(3 + x)^2 \sqrt{3 - x^2}}$

$= \frac{-3x - x^2 - 3 + x^2}{(3 + x)^2 \sqrt{3 - x^2}}$

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

Thus, the derivative is:

$\frac{d}{dx} \left( \frac{\sqrt{3 - x^2}}{3 + x} \right) = \frac{-3(x + 1)}{(3 + x)^2 \sqrt{3 - x^2}}$

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Do you want further details or have any questions?

Here are five related questions to expand your understanding:

1. How do you apply the product rule and chain rule together in derivatives?
2. What is the geometrical interpretation of the derivative of a function?
3. How does the quotient rule compare to the product rule?
4. Can you find the second derivative of the same function?
5. How would this derivative behave as $x$ approaches 3?

Tip: Always simplify expressions before applying derivative rules to make the process easier.