The function is given as:

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

To find the derivative, $f'(x)$, we'll use the quotient rule, which is:

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

Step 1: Identify $u(x)$ and $v(x)$

• $u(x) = 3x^3$
• $v(x) = 2x^3 + 1$

Step 2: Find the derivatives $u'(x)$ and $v'(x)$

• $u'(x) = \frac{d}{dx}(3x^3) = 9x^2$
• $v'(x) = \frac{d}{dx}(2x^3 + 1) = 6x^2$

Step 3: Apply the quotient rule

Now, applying the quotient rule:

$f'(x) = \frac{(9x^2)(2x^3 + 1) - (3x^3)(6x^2)}{(2x^3 + 1)^2}$

Step 4: Simplify the numerator

Expand the terms in the numerator:

$f'(x) = \frac{9x^2(2x^3 + 1) - 18x^5}{(2x^3 + 1)^2}$

Distribute the $9x^2$ in the first term:

$f'(x) = \frac{18x^5 + 9x^2 - 18x^5}{(2x^3 + 1)^2}$

The $18x^5$ terms cancel out, leaving:

$f'(x) = \frac{9x^2}{(2x^3 + 1)^2}$

Thus, the derivative in simplified form is:

$f'(x) = \frac{9x^2}{(2x^3 + 1)^2}$

Would you like a detailed explanation of any step, or to clarify any part of the process?

Here are 5 related questions to practice:

1. What is the derivative of $\frac{x^4}{3x^2 + 2}$?
2. How do you apply the quotient rule when the numerator is a constant?
3. What happens to the derivative when both numerator and denominator have the same degree?
4. Can the chain rule be applied to solve this function instead of the quotient rule?
5. How does the quotient rule apply when both terms are linear?

Tip: Always simplify the numerator and denominator before applying the quotient rule to make calculations easier!