Problem 3 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(t)=(t-1)(2 t+1)\)

Step-by-Step Solution

Answer

The derivative of the function \(f(t) = (t-1)(2t+1)\) is \(f'(t) = 4t - 1\).

Step 1: Find the derivative of g(t)

We have \(g(t) = t - 1\). To find the derivative, we will differentiate with respect to \(t\): \(g'(t) = \frac{d}{dt} (t - 1)\) Using basic differentiation rules, we get: \(g'(t) = 1 - 0 = 1\)

Step 2: Find the derivative of h(t)

We have \(h(t) = 2t + 1\). To find the derivative, we will differentiate with respect to \(t\): \(h'(t) = \frac{d}{dt} (2t + 1)\) Using basic differentiation rules, we get: \(h'(t) = 2 + 0 = 2\)

Step 3: Apply the Product Rule to find f'(t)

Now that we have \(g'(t)\) and \(h'(t)\), we can apply the Product Rule formula: \(f'(t) = g'(t) \cdot h(t) + g(t) \cdot h'(t)\) Substitute the values of \(g'(t)\), \(g(t)\), \(h'(t)\), and \(h(t)\), we get: \(f'(t) = (1) \cdot (2t + 1) + (t - 1) \cdot (2)\) Now simplify the expression: \(f'(t) = 2t + 1 + 2t - 2\)

Step 4: Combine like terms

Combine the like terms to obtain the final expression for the derivative: \(f'(t) = (2t + 2t) + (1 - 2) \) So, the derivative of the function is: \(f'(t) = 4t - 1\) The derivative of the function \(f(t) = (t-1)(2t+1)\) is \(f'(t) = 4t - 1\).

Key Concepts

Product Rule

The Product Rule in differentiation is a fundamental concept. It helps us find the derivative of two functions that are multiplied together. If you have a function that is the product of two simpler functions, say \( u(t) \) and \( v(t) \), the Product Rule states:

• \( (uv)' = u'v + uv' \)

This means that the derivative of \( u \) times \( v \) is \( u' \) (the derivative of \( u \)) times \( v \), plus \( u \) times \( v' \) (the derivative of \( v \)). For our function \( f(t) = (t-1)(2t+1) \), we treated \( (t-1) \) as \( g(t) \) and \( (2t+1) \) as \( h(t) \). The step-by-step process followed this rule to calculate \( f'(t) = g'(t) \cdot h(t) + g(t) \cdot h'(t) \). This approach simplifies the task of finding derivatives for products of functions.

Basic Differentiation Rules

Basic differentiation rules are key for finding derivatives easily. These rules include the power rule, constant rule, and basic linearity, among others. For instance, the derivative of a constant is zero. The derivative of a basic term like \( t \) is straightforward.

• For \( g(t) = t - 1 \), using the rule that the derivative of \( t \) is 1, we find \( g'(t) = 1 \).
• If you have \( h(t) = 2t + 1 \), employing the rule that the derivative of \( at \) is \( a \), we deduce \( h'(t) = 2 \).

Knowing and applying these basic rules correctly saves time and makes the differentiation process much easier. Such fundamental rules are usually the first things students learn when they start with calculus, laying groundwork for more complicated concepts.

Finding Derivatives

Finding derivatives involves applying differentiation rules to find the rate at which a function is changing. In calculus, this is crucial because derivatives tell us about the slope of the function at any given point.

• In the exercise, we first found \( g'(t) = 1 \) and \( h'(t) = 2 \), using basic differentiation rules.
• Next, we applied the Product Rule, which requires using these derivatives to find \( f'(t) \).

By simplifying and combining like terms, such as \( 2t + 2t \) and \( 1 - 2 \), we determined the final derivative as \( f'(t) = 4t - 1 \). This demonstrates how understanding and applying rules methodically allows finding derivatives for even slightly more complex functions like products. The skill of derivative finding is key to understanding curves and solving real-world problems involving rates of change.