To differentiate the function $f(x) = 3x^{2/3}$ using first principles, we start by applying the definition of the derivative:

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

Step 1: Substituting the function into the definition of the derivative

The given function is $f(x) = 3x^{2/3}$. So,

$f(x+h) = 3(x+h)^{2/3}$

Now substitute $f(x+h)$ and $f(x)$ into the definition of the derivative:

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

Step 2: Factor using the difference of cubes

The key to simplifying the expression involves using the difference of cubes. Notice that $a^{2/3} - b^{2/3}$ can be factored as a cube root difference:

$a^{2/3} - b^{2/3} = \frac{a - b}{a^{1/3} + a^{2/3}b^{1/3} + b^{1/3}}$

Here, let $a = (x+h)^{2/3}$ and $b = x^{2/3}$.

First, factor out the 3:

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

Now use the difference of cubes form:

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

Step 3: Simplify

Simplify the numerator:

$x+h - x = h$

So the expression becomes:

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

Cancel $h$ from the numerator and denominator:

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

Step 4: Take the limit as $h \to 0$

As $h \to 0$, $(x+h)^{1/3} \to x^{1/3}$ and $(x+h)^{2/3} \to x^{2/3}$. Therefore, the expression simplifies to:

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

Finally, simplify:

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

Conclusion:

The derivative of $f(x) = 3x^{2/3}$ using first principles is:

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

Would you like more details or have any questions about the process?

Here are 5 questions you might explore next:

1. How does the difference of cubes factorization generally work in calculus?
2. What is the geometric interpretation of the derivative found?
3. Can we apply this method to higher powers of $x$?
4. What is the relationship between first principles and standard differentiation rules?
5. How can we use first principles to differentiate logarithmic functions?

Tip: Always try to recognize patterns, such as difference of cubes, when simplifying expressions in calculus!