Math Problem Statement

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

Solution

To differentiate the function f(x)=3x34f(x) = \frac{3x^3}{4} using first principles, we begin with the definition of the derivative:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Given the function f(x)=3x34f(x) = \frac{3x^3}{4}, we substitute into the definition:

f(x)=limh03(x+h)343x34hf'(x) = \lim_{h \to 0} \frac{\frac{3(x+h)^3}{4} - \frac{3x^3}{4}}{h}

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

(x+h)3=x3+3x2h+3xh2+h3(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3

Substitute this into the expression for the derivative:

f(x)=limh03(x3+3x2h+3xh2+h3)43x34hf'(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)=limh03x3+9x2h+9xh2+3h343x34hf'(x) = \lim_{h \to 0} \frac{\frac{3x^3 + 9x^2h + 9xh^2 + 3h^3}{4} - \frac{3x^3}{4}}{h}

f(x)=limh03x3+9x2h+9xh2+3h33x34hf'(x) = \lim_{h \to 0} \frac{\frac{3x^3 + 9x^2h + 9xh^2 + 3h^3 - 3x^3}{4}}{h}

f(x)=limh09x2h+9xh2+3h34hf'(x) = \lim_{h \to 0} \frac{\frac{9x^2h + 9xh^2 + 3h^3}{4}}{h}

Now, factor hh out from the numerator:

f(x)=limh0h(9x2+9xh+3h24)hf'(x) = \lim_{h \to 0} \frac{h\left(\frac{9x^2 + 9xh + 3h^2}{4}\right)}{h}

Since hh is in both the numerator and the denominator, we can cancel it out:

f(x)=limh09x2+9xh+3h24f'(x) = \lim_{h \to 0} \frac{9x^2 + 9xh + 3h^2}{4}

Now, take the limit as hh approaches 0:

f(x)=9x2+9x(0)+3(0)24=9x24f'(x) = \frac{9x^2 + 9x(0) + 3(0)^2}{4} = \frac{9x^2}{4}

Therefore, the derivative of the function f(x)=3x34f(x) = \frac{3x^3}{4} is:

f(x)=9x24f'(x) = \frac{9x^2}{4}

Summary:

  • We expanded (x+h)3(x+h)^3 using the binomial theorem.
  • Simplified the expression using algebraic manipulation.
  • Applied the limit as hh 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)=x3g(x) = x^3 using first principles?
  2. What is the derivative of f(x)=5x42f(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.

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Math Problem Analysis

Mathematical Concepts

Differentiation
First Principles
Binomial Theorem
Difference of Cubes

Formulas

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

Theorems

Binomial Theorem

Suitable Grade Level

Grades 11-12