Problem 5 - Calculus Concepts

Question

Use the limit definition of the derivative (algebraic method) to confirm the statements. The derivative of \(f(x)=x^{3}\) is \(f^{\prime}(x)=3 x^{2}\).

Step-by-Step Solution

Answer

The derivative of \(f(x)=x^3\) is \(f'(x)=3x^2\), confirmed by the limit definition.

Step 1: Understand the Limit Definition of the Derivative

The limit definition of the derivative for a function \(f(x)\) is given by:\[f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\]This formula allows us to find the derivative of a function at a given point \(x\).

Step 2: Substitute the Function into the Definition

Substitute \(f(x) = x^3\) into the limit definition:\[f'(x) = \lim_{{h \to 0}} \frac{(x+h)^3 - x^3}{h}\]

Step 3: Expand the Cubic Expression

Expand \((x+h)^3\) using the binomial theorem:\[(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\]

Step 4: Simplify the Expression

Subtract \(x^3\) from the expanded form:\[(x+h)^3 - x^3 = 3x^2h + 3xh^2 + h^3\]Now substitute this back into the limit definition:\[f'(x) = \lim_{{h \to 0}} \frac{3x^2h + 3xh^2 + h^3}{h}\]

Step 5: Factor Out \(h\) and Simplify

Factor \(h\) out of the numerator:\[f'(x) = \lim_{{h \to 0}} \frac{h(3x^2 + 3xh + h^2)}{h}\]Cancel \(h\) from the numerator and denominator:\[f'(x) = \lim_{{h \to 0}} (3x^2 + 3xh + h^2)\]

Step 6: Evaluate the Limit

As \(h\) approaches 0, the terms involving \(h\) vanish, so:\[f'(x) = 3x^2 + 3x(0) + 0^2 = 3x^2\]Thus, the derivative of \(x^3\) is \(3x^2\).

Key Concepts

Derivative Calculation

The concept of derivative calculation is crucial in understanding how functions change. To calculate the derivative, we use the **limit definition** which captures the idea of finding the slope of the tangent line to a curve at any point. For a function like \(f(x) = x^3\), the derivative is found by evaluating the limit:

• Take the difference in function values: \(f(x + h) - f(x)\)
• Divide by \(h\)
• Take the limit as \(h\) approaches 0

By substituting the given function into the limit formula, we find \(f'(x) = \lim_{{h \to 0}} \frac{(x+h)^3 - x^3}{h}\). This process involves expanding expressions and simplifying, resulting in a clean derivative formula that shows how the function behaves around a point. This powerful technique is foundational in calculus.

Binomial Theorem

The **binomial theorem** helps in expanding expressions of the form \((x+h)^n\). It's essential when dealing with polynomials in derivative calculations. For example, to expand \((x+h)^3\), we use the binomial pattern:

• \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\)
• This expansion simplifies calculations by expressing a complex term in manageable parts.

The binomial theorem states that \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). Here, it's a perfect fit for polynomial functions as it breaks down the complexity of terms deriving from \(x\) and \(h\). This simplification is key to more easily applying the limit definition of a derivative.

Algebraic Method

The **algebraic method** is used to simplify expressions when calculating derivatives. Starting with the limit definition, which is an algebraic expression itself, involves substitution and simplification:

• We first substitute \(f(x) = x^3\) into the limit definition.
• Utilize tools like the binomial theorem for expansion.
• Simplify expressions by canceling and factoring terms.

Factoring allows for reducing terms, such as taking \(h\) out of \(3x^2h + 3xh^2 + h^3\) to \(h(3x^2 + 3xh + h^2)\), and then simplifying further by canceling \(h\). Once simplified, evaluating the limit is easier because you're often left with straightforward polynomial terms as \(h\) approaches zero. This algebraic manipulation makes finding derivatives accurate and efficient.