Using first principles, differentiate �(�) = 3�!⁄". Show your work. [Hint: use the
‘difference of cubes.’]

To differentiate the function \( f(x) = \frac{3x^3}{4} \) using first principles, we begin with the definition of the derivative:

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

Given the function \( f(x) = \frac{3x^3}{4} \), we substitute into the definition:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{3(x+h)^3}{4} - \frac{3x^3}{4}}{h}
\]

First, let's expand \( (x+h)^3 \) using the binomial theorem:

\[
(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
\]

Substitute this into the expression for the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{3(x^3 + 3x^2h + 3xh^2 + h^3)}{4} - \frac{3x^3}{4}}{h}
\]

Simplify the expression:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{3x^3 + 9x^2h + 9xh^2 + 3h^3}{4} - \frac{3x^3}{4}}{h}
\]

\[
f'(x) = \lim_{h \to 0} \frac{\frac{3x^3 + 9x^2h + 9xh^2 + 3h^3 - 3x^3}{4}}{h}
\]

\[
f'(x) = \lim_{h \to 0} \frac{\frac{9x^2h + 9xh^2 + 3h^3}{4}}{h}
\]

Now, factor \( h \) out from the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{h\left(\frac{9x^2 + 9xh + 3h^2}{4}\right)}{h}
\]

Since \( h \) is in both the numerator and the denominator, we can cancel it out:

\[
f'(x) = \lim_{h \to 0} \frac{9x^2 + 9xh + 3h^2}{4}
\]

Now, take the limit as \( h \) approaches 0:

\[
f'(x) = \frac{9x^2 + 9x(0) + 3(0)^2}{4} = \frac{9x^2}{4}
\]

Therefore, the derivative of the function \( f(x) = \frac{3x^3}{4} \) is:

\[
f'(x) = \frac{9x^2}{4}
\]

**Summary**:
- We expanded \( (x+h)^3 \) using the binomial theorem.
- Simplified the expression using algebraic manipulation.
- Applied the limit as \( h \) approaches 0 to obtain the final derivative.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. How would you differentiate \( g(x) = x^3 \) using first principles?
2. What is the derivative of \( f(x) = \frac{5x^4}{2} \) using first principles?
3. Can you explain the difference of cubes formula and how it applies here?
4. What is the general rule for differentiating a cubic polynomial?
5. How does using first principles differ from using standard differentiation rules?

**Tip**: Understanding first principles is fundamental as it forms the basis of all differentiation techniques.

Using first principles, differentiate \(f(x) = \frac{3x^3}{4}\). Show your work. [Hint: use the ‘difference of cubes.’]

Explore the process of differentiating \(f(x) = \frac{3x^3}{4}\) using first principles. This detailed solution includes the application of the binomial theorem and the difference of cubes, making it suitable for advanced high school students (Grades 11-12).

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