To find the derivative of the function $f(x) = \frac{3x^3}{x^2 - 4}$, we can use the quotient rule. The quotient rule states that if we have a function $f(x) = \frac{g(x)}{h(x)}$, then

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

where $g(x) = 3x^3$ and $h(x) = x^2 - 4$.

Step 1: Differentiate $g(x)$ and $h(x)$

1. $g(x) = 3x^3$
   $g'(x) = 9x^2$

2. $h(x) = x^2 - 4$
   $h'(x) = 2x$

Step 2: Apply the Quotient Rule

Now, using the quotient rule:

$f'(x) = \frac{(9x^2)(x^2 - 4) - (3x^3)(2x)}{(x^2 - 4)^2}$

Step 3: Simplify the Expression

Let's simplify the terms in the numerator:

1. Expanding $(9x^2)(x^2 - 4)$: $9x^2 \cdot x^2 - 9x^2 \cdot 4 = 9x^4 - 36x^2$

2. Expanding $(3x^3)(2x)$: $3x^3 \cdot 2x = 6x^4$

So, the numerator becomes:

$9x^4 - 36x^2 - 6x^4 = 3x^4 - 36x^2$

Therefore, we have:

$f'(x) = \frac{3x^4 - 36x^2}{(x^2 - 4)^2}$

Final Answer

$f'(x) = \frac{3x^2(x^2 - 12)}{(x^2 - 4)^2}$

Would you like further clarification on any steps?

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1. How is the quotient rule applied to complex functions?
2. What happens to $f'(x)$ if $x = 2$ or $x = -2$?
3. Could the function be simplified before differentiation?
4. What is the importance of simplification in calculus?
5. How do we handle higher-order derivatives of rational functions?

Tip: Simplifying expressions before applying differentiation rules can sometimes make calculations much easier.