Math Problem Statement

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

Solution

To differentiate f(x)=3x23f(x) = 3x^{\frac{2}{3}} using first principles, we use 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}

Step 1: Substitute f(x)=3x23f(x) = 3x^{\frac{2}{3}} into the definition

f(x)=limh03(x+h)233x23hf'(x) = \lim_{h \to 0} \frac{3(x+h)^{\frac{2}{3}} - 3x^{\frac{2}{3}}}{h}

We can factor out the constant 33:

f(x)=3limh0(x+h)23x23hf'(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:

a3b3=(ab)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

To use this hint, we need to rewrite (x+h)23x23(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:

a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b)

Here, a=(x+h)13a = (x+h)^{\frac{1}{3}} and b=x13b = 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 ((x+h)13x13)\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:

ab=a3b3a2+ab+b2a - b = \frac{a^3 - b^3}{a^2 + ab + b^2}

Let’s apply this formula with a=(x+h)13a = (x+h)^{\frac{1}{3}} and b=x13b = x^{\frac{1}{3}}:

(x+h)13x13=h(x+h)23+(x+h)13x13+x23(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)=3limh0h(x+h)23+(x+h)13x13+x23[(x+h)13+x13]hf'(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 hh terms cancel out:

f(x)=3limh0(x+h)13+x13(x+h)23+(x+h)13x13+x23f'(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 h0h \to 0

As h0h \to 0, (x+h)13x13(x+h)^{\frac{1}{3}} \to x^{\frac{1}{3}}, and the expression simplifies:

f(x)=32x133x23=2x13f'(x) = 3 \cdot \frac{2x^{\frac{1}{3}}}{3x^{\frac{2}{3}}} = 2x^{-\frac{1}{3}}

Final Answer:

Thus, the derivative of f(x)=3x23f(x) = 3x^{\frac{2}{3}} is:

f(x)=2x13f'(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 xnx^{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 xkx^{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.

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Math Problem Analysis

Mathematical Concepts

Differentiation
First Principles
Algebra
Difference of Cubes

Formulas

f'(x) = lim(h -> 0) [f(x+h) - f(x)] / h
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a^2 - b^2 = (a - b)(a + b)

Theorems

Difference of Cubes
Limit Definition of the Derivative

Suitable Grade Level

Grades 11-12 (Advanced Calculus)