Problem 2 - Calculus

Question

Use the definition $$f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$$ to find the indicated derivative. \(f^{\prime}(2)\) if \(f(t)=(2 t)^{2}\)

Step-by-Step Solution

Answer

The derivative \( f'(2) \) is 16.

Step 1: Understanding the Problem

We need to find the derivative of the function \( f(t) = (2t)^2 \) at \( t = 2 \). The derivative, \( f'(t) \), is calculated using the limit definition: \( f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \). We will apply this definition to find \( f'(2) \).

Step 2: Substituting into the Function

Start by substituting \( t + h \) into the function \( f(t) = (2t)^2 \). This gives: \[ f(t + h) = (2(t + h))^2 = (2t + 2h)^2 \].

Step 3: Expanding \( f(t+h) \)

Expand the expression \((2t + 2h)^2\):\[ (2t + 2h)^2 = (2t)^2 + 2(2t)(2h) + (2h)^2 = 4t^2 + 8th + 4h^2 \].

Step 4: Compute the Difference \( f(t+h) - f(t) \)

Substitute \( f(t) = (2t)^2 = 4t^2 \) and \( f(t+h) \) from step 3 into the difference:\[ f(t+h) - f(t) = (4t^2 + 8th + 4h^2) - 4t^2 = 8th + 4h^2 \].

Step 5: Form the Difference Quotient

Set up the difference quotient: \[ \frac{f(t+h) - f(t)}{h} = \frac{8th + 4h^2}{h} = 8t + 4h \], by dividing each term by \( h \).

Step 6: Taking the Limit as \( h \) Approaches Zero

Take the limit of the difference quotient as \( h \to 0 \):\[ \lim_{h \to 0} (8t + 4h) = 8t \], because \( 4h \) approaches zero as \( h \to 0 \).

Step 7: Substitute \( t = 2 \) to Find \( f'(2) \)

Substitute \( t = 2 \) into the result from step 6 to find the derivative at this point:\[ f'(2) = 8 \times 2 = 16 \].

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It provides a precise way to calculate the rate of change, or slope, of a function at a specific point. The official definition is expressed as:\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}, \]where \( f'(c) \) represents the derivative of the function \( f \) at the point \( c \). This equation captures how the function \( f \) changes as \( h \), a small increment, approaches zero.
Understanding this definition involves several parts:

• \( f(c+h) \): This term represents the function value at a small distance \( h \) away from \( c \).
• \( f(c) \): This represents the function value at the exact point \( c \).
• \( \frac{f(c+h) - f(c)}{h} \): This quotient calculates the average rate of change over the interval from \( c \) to \( c+h \).

The limit, \( \lim_{h \to 0} \), refines this average rate of change to an instantaneous rate of change, revealing the exact slope at point \( c \). This method forms the backbone of learning about derivatives before moving to shortcuts like rules of differentiation.

Polynomial Derivatives

Polynomials are common mathematical expressions that are relatively straightforward to differentiate due to their consistent structure. A polynomial function is represented as a sum of terms, each of the form \( a_n t^n \) where \( a_n \) is a coefficient and \( n \) is a non-negative integer.
When finding derivatives of polynomial functions, each term is differentiated using the power rule. The power rule states that the derivative of \( t^n \) is \( n \cdot t^{n-1} \). For example, the derivative of \( (2t)^2 \) can be expanded to \( 4t^2 \) so that we can apply the power rule effectively.
To differentiate a complete polynomial, perform the power rule for each term:

• Derive each term separately.
• Apply any necessary constants or coefficients.
• Sum the derivatives of the individual terms to find the derivative of the entire polynomial.

This systematic approach provides a simple but powerful method to handle polynomial derivatives, making it an essential tool for solving calculus problems.

Calculating Derivatives at a Point

When solving problems, knowing how to calculate a derivative at a specific point is important, especially in understanding the behavior of functions in calculus. Here’s how to approach it:

• Understand the Function: First, identify the function you are interested in. For this example, it's \( f(t) = (2t)^2 \).
• Use the Limit Definition: As seen previously, start with the limit definition of a derivative. This provides a detailed path to finding \( f'(t) \) by directly considering the function's behavior as \( h \) approaches zero.
• Simplify Terms: Simplify any expressions before taking the limit, such as the "difference quotient" \( \frac{f(t+h)-f(t)}{h} \).
• Substitute the Point: Once the general derivative \( f'(t) \) is found, substitute the specific value of \( t \) to find the derivative at that point.

This method provides a deep understanding of the derivative's meaning and is crucial for more complex problem-solving in calculus. With practice, shifting from general derivatives to specific evaluations becomes second nature.