Problem 4 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {f(x)=\left(x^{2}+1\right)(2 x+5)} & {(-1,6)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point (-1,6) is -2. The product rule was used to find the derivative.

Step 1: Identify the different functions

We have two different functions in the given function, defined as \(f(x) = (x^2 + 1)(2x + 5)\). This can be rewritten as \( u(x) = x^2 + 1 \) and \( v(x) = 2x + 5 \) where \( u(x) \) and \( v(x) \) are individual functions.

Step 2: Apply the product rule

The product rule states that the derivative of a product of functions is given by \( (u.v)' = u'.v + u.v' \). Apply this rule to find the derivative of \( f(x) \), which is written as \( f'(x) \). Thus, \( f'(x) = u'(x).v(x) + u(x).v'(x) \). To do this, we need to find the derivatives of \( u(x) \) and \( v(x) \), namely \( u'(x) \) and \( v'(x) \). The derivative \( u'(x) = 2x \) and \( v'(x) = 2 \). Substitute these into the product rule to get \( f'(x) = 2x(2x + 5) + (x^2 + 1)2 \).

Step 3: Simplify the derivative

Simplify the expression to find a simpler form of the derivative, which gives: \( f'(x) = 4x^2 + 10x + 2x^2 + 2 = 6x^2 + 10x + 2 \).

Step 4: Evaluate the derivative at the given point

The point given is (-1,6). The x-coordinate is what's important here, because we want to find the derivative at this specific x-value. Plugging \( x = -1 \) into the derivative yields \( f'(-1) = 6(-1)^2 + 10(-1) + 2 = 6 - 10 + 2 = -2 \)

Key Concepts

Understanding the Product Rule

When working with calculus and examining the derivative of a function, it's essential to understand the product rule. Imagine you are asked to differentiate a function that is formed by multiplying two simpler functions, like in the exercise, where we have a function written as

• \((x^2 + 1)(2x + 5)\)

The product rule aids us by providing a formula for finding the derivative of such products of functions.
The product rule formula is:

• \((u \, v)' = u' \, v + u \, v'\)

Here, \(u\) and \(v\) are two functions, and their derivatives are \(u'\) and \(v'\), respectively. Following the formula, you multiply the derivative of the first function by the second function and add it to the product of the first function and the derivative of the second function. This approach can be invaluable for correctly tackling derivatives in mathematical differentiation.

The Role of Calculus

In mathematical differentiation, calculus plays a significant role. Calculus is the branch of mathematics that studies change, just like in this exercise where we determine how a function changes at a specific point.
Derivatives are a foundational concept within calculus, representing the rate of change of a function as its input changes.

• When you take a derivative, you are essentially finding how steep a curve is at particular points.

Practicing calculus helps you understand various physical systems, model real-world scenarios, and predict change's behavior.
In this way, calculus is not just about solving equations—it is about understanding change and motion, key elements in mathematics and physics.

Mathematical Differentiation Basics

Differentiation is a fundamental operation in calculus, and understanding it is crucial for solving many problems involving rates of change. Differentiating a function essentially means finding its derivative to determine how the function behaves as the input changes.
In the context of the exercise, we've broken down the original function into simpler components, allowing us to individually differentiate them and then apply rules like the product rule.

• Differentiation follows specific rules and principles, ensuring that we handle the derivatives correctly.
• Key principles include constant, power, sum, and product rules.

Using these rules systematically allows us to tackle complex functions effectively, translating to more accurate mathematical modeling and predictions. Hence, understanding these basics can make mastering differentiation more manageable and rewarding.