To differentiate $f(x) = 3x^{2/3}$ using first principles, we need to apply the definition of the derivative. The derivative of a function $f(x)$ is given by:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 1: Apply the definition of the derivative

Here, $f(x) = 3x^{2/3}$, so the derivative is:

$f'(x) = \lim_{h \to 0} \frac{3(x+h)^{2/3} - 3x^{2/3}}{h}$

Factor out the 3 to simplify:

$f'(x) = 3 \lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{h}$

Step 2: Using the Hint (Difference of Cubes)

We know that the expression $a^{2/3} - b^{2/3}$ can be treated similarly to the difference of cubes formula. The difference of cubes formula is:

$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

To apply a similar idea to fractional powers like $x^{2/3}$, we write the difference as:

$(x+h)^{2/3} - x^{2/3} = \frac{\left[(x+h)^{2} - x^2\right]}{(x+h)^{1/3} + x^{1/3}}$

Step 3: Expanding the Numerator

Now expand the numerator $(x+h)^2 - x^2$:

$(x+h)^2 = x^2 + 2xh + h^2$ $(x+h)^2 - x^2 = 2xh + h^2$

Thus, we have:

$(x+h)^{2/3} - x^{2/3} = \frac{2xh + h^2}{(x+h)^{1/3} + x^{1/3}}$

Step 4: Substitute into the Derivative

Substitute this into the derivative expression:

$f'(x) = 3 \lim_{h \to 0} \frac{\frac{2xh + h^2}{(x+h)^{1/3} + x^{1/3}}}{h}$

This simplifies to:

$f'(x) = 3 \lim_{h \to 0} \frac{2xh + h^2}{h \left( (x+h)^{1/3} + x^{1/3} \right)}$

Step 5: Simplify the Expression

Factor out $h$ from the numerator:

$f'(x) = 3 \lim_{h \to 0} \frac{h(2x + h)}{h \left( (x+h)^{1/3} + x^{1/3} \right)}$

Cancel the $h$ terms:

$f'(x) = 3 \lim_{h \to 0} \frac{2x + h}{(x+h)^{1/3} + x^{1/3}}$

As $h \to 0$, the term $2x + h$ approaches $2x$, and $(x+h)^{1/3}$ approaches $x^{1/3}$. Thus, we get:

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

Final Answer:

$f'(x) = \frac{3}{x^{1/3}}$

Summary:

We have used first principles and the hint involving the difference of cubes to find the derivative of $f(x) = 3x^{2/3}$. The final result is:

$f'(x) = \frac{3}{x^{1/3}}$

---

Would you like further details or clarification? Here are some additional questions you could explore:

1. How does this method compare to the power rule for differentiation?
2. What is the geometric interpretation of $f'(x) = \frac{3}{x^{1/3}}$?
3. How would the derivative change if the exponent were $\frac{5}{3}$ instead of $\frac{2}{3}$?
4. Can this method be applied to find the second derivative of $f(x) = 3x^{2/3}$?
5. How would you interpret the derivative when $x = 0$?

Tip: The difference of cubes formula is a powerful tool in simplifying complex algebraic expressions, especially when dealing with roots or fractional powers.