Using first principles, differentiate f(x) = 3x^(2/3). Show your work. [Hint: use the ‘difference of cubes.’]

To differentiate \( f(x) = 3x^{\frac{2}{3}} \) using **first principles**, we use the definition of the derivative:

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

### Step 1: Substitute \( f(x) = 3x^{\frac{2}{3}} \) into the definition

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

We can factor out the constant \( 3 \):

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

### Step 2: Use the **difference of cubes** formula

We know that the difference of cubes can be written as:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

To use this hint, we need to rewrite \( (x+h)^{\frac{2}{3}} - x^{\frac{2}{3}} \) as a difference of cubes. Let’s rewrite the exponents in a cubic form:

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

Now apply the difference of cubes identity:

\[
a^2 - b^2 = (a - b)(a + b)
\]

Here, \( a = (x+h)^{\frac{1}{3}} \) and \( b = x^{\frac{1}{3}} \), so:

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

### Step 3: Substitute back into the limit expression

Now, substitute this back into the limit:

\[
f'(x) = 3 \lim_{h \to 0} \frac{\left[ (x+h)^{\frac{1}{3}} - x^{\frac{1}{3}} \right] \cdot \left[ (x+h)^{\frac{1}{3}} + x^{\frac{1}{3}} \right]}{h}
\]

### Step 4: Focus on simplifying the first factor

Now, consider only the first factor \( \left( (x+h)^{\frac{1}{3}} - x^{\frac{1}{3}} \right) \). This expression is a difference of two cube roots, and we can simplify it using the **difference of cubes** identity:

\[
a - b = \frac{a^3 - b^3}{a^2 + ab + b^2}
\]

Let’s apply this formula with \( a = (x+h)^{\frac{1}{3}} \) and \( b = x^{\frac{1}{3}} \):

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

### Step 5: Substitute this into the limit expression

Now substitute this result back into the limit:

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

The \( h \) terms cancel out:

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

### Step 6: Take the limit as \( h \to 0 \)

As \( h \to 0 \), \( (x+h)^{\frac{1}{3}} \to x^{\frac{1}{3}} \), and the expression simplifies:

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

### Final Answer:

Thus, the derivative of \( f(x) = 3x^{\frac{2}{3}} \) is:

\[
f'(x) = 2x^{-\frac{1}{3}}
\]

Would you like more details or have any questions?

Here are five related questions you might find useful:
1. How do you apply the difference of cubes in other situations?
2. How can you use first principles to differentiate \(x^{n}\) in general?
3. Can you generalize the first-principles method for fractional powers?
4. How does this compare to using standard derivative rules for power functions?
5. What happens if the exponent in \(x^{k}\) is negative or irrational?

**Tip:** When differentiating using first principles, breaking down complex terms (like roots or powers) using algebraic identities like the difference of cubes simplifies the process.

Learn how to differentiate f(x) = 3x^(2/3) using first principles and the difference of cubes identity. This solution breaks down the process step-by-step, using algebraic techniques and the limit definition of the derivative. Suitable for advanced calculus students in Grades 11-12.

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