Find Derivative Of F(x) = (2x² - 5)(4x + 4) And Evaluate F'(5)

Jul 16, 2025 by ADMIN 63 views

Finding the Derivative and Evaluating at a Point If f(x) = (2x² - 5)(4x + 4)

In calculus, finding the derivative of a function is a fundamental operation. The derivative, denoted as f'(x), represents the instantaneous rate of change of the function f(x) with respect to its input variable x. It provides valuable insights into the function's behavior, such as its slope, increasing or decreasing intervals, and concavity. This article delves into the process of finding the derivative of a given function and evaluating it at a specific point. We will use the product rule and power rule of differentiation to find f'(x) when f(x) = (2x² - 5)(4x + 4) and then determine the value of f'(5).

Understanding the Problem

The problem at hand requires us to perform two key tasks: first, to find the derivative of the function f(x) = (2x² - 5)(4x + 4), and second, to evaluate this derivative at the point x = 5. The given function is a product of two expressions, namely (2x² - 5) and (4x + 4). Therefore, we will need to apply the product rule of differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In mathematical notation, if we have two functions u(x) and v(x), the product rule can be expressed as:

(d/dx)[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)

In our case, we can identify u(x) = 2x² - 5 and v(x) = 4x + 4. To apply the product rule, we first need to find the derivatives of these individual functions. For this, we will use the power rule of differentiation, which states that the derivative of x^n is nx^(n-1), where n is any real number. We also need to remember that the derivative of a constant is zero.

After finding the derivative f'(x), we will substitute x = 5 into the expression to find the value of f'(5). This will give us the instantaneous rate of change of the function at the specific point x = 5. This process is fundamental in various applications of calculus, including optimization problems, curve sketching, and understanding rates of change in physical systems.

Step-by-Step Solution

1. Identify u(x) and v(x)

As mentioned earlier, we identify the two functions that are being multiplied together:

u(x) = 2x² - 5
v(x) = 4x + 4

These two functions are the building blocks for applying the product rule, and correctly identifying them is crucial for the next steps in finding the derivative.

2. Find u'(x) and v'(x)

Now, we need to find the derivatives of u(x) and v(x) separately. This involves applying the power rule and the constant multiple rule of differentiation. The power rule, as a reminder, states that the derivative of x^n is nx^(n-1), and the constant multiple rule states that the derivative of a constant times a function is the constant times the derivative of the function. Additionally, the derivative of a constant term is zero.

Let's find u'(x), the derivative of u(x) = 2x² - 5:

The derivative of 2x² is 2 * 2x^(2-1) = 4x (using the power rule and constant multiple rule).
The derivative of -5 is 0 (since it's a constant).
Therefore, u'(x) = 4x + 0 = 4x.

Next, let's find v'(x), the derivative of v(x) = 4x + 4:

The derivative of 4x is 4 * 1x^(1-1) = 4 (using the power rule and constant multiple rule).
The derivative of 4 is 0 (since it's a constant).
Therefore, v'(x) = 4 + 0 = 4.

Having found u'(x) = 4x and v'(x) = 4, we are now prepared to apply the product rule.

3. Apply the Product Rule

The product rule states that f'(x) = u'(x)v(x) + u(x)v'(x). Now we substitute the expressions we found in the previous steps:

f'(x) = (4x)(4x + 4) + (2x² - 5)(4)

This step is the core of finding the derivative of the original function. By correctly applying the product rule and substituting the derivatives of the individual functions, we have an expression for f'(x).

4. Simplify f'(x)

To simplify the expression, we need to expand the terms and combine like terms:

f'(x) = 16x² + 16x + 8x² - 20

Now, we combine the terms with the same power of x:

f'(x) = (16x² + 8x²) + 16x - 20

f'(x) = 24x² + 16x - 20

This simplified expression for f'(x) is a quadratic function, which represents the instantaneous rate of change of the original function f(x) at any point x. This simplification makes it easier to evaluate f'(x) at specific points, such as x = 5, which is our next step.

5. Evaluate f'(5)

To find f'(5), we substitute x = 5 into the simplified expression for f'(x):

f'(5) = 24(5)² + 16(5) - 20

First, we calculate 5 squared, which is 25:

f'(5) = 24(25) + 16(5) - 20

Next, we perform the multiplications:

f'(5) = 600 + 80 - 20

Finally, we add and subtract the terms:

f'(5) = 680 - 20

f'(5) = 660

Therefore, the value of the derivative of the function f(x) at x = 5 is 660. This value represents the slope of the tangent line to the graph of f(x) at the point where x = 5. It also indicates the instantaneous rate of change of the function at that specific point. This result is a numerical answer to the second part of the original problem, where we were asked to evaluate the derivative at x = 5.

Summary and Conclusion

In this comprehensive exploration, we successfully found the derivative of the function f(x) = (2x² - 5)(4x + 4) and evaluated it at x = 5. The process involved several key steps:

Identifying the two functions u(x) and v(x) that form the product in f(x).
Finding the derivatives u'(x) and v'(x) using the power rule and constant multiple rule.
Applying the product rule to find f'(x).
Simplifying the expression for f'(x) by expanding and combining like terms.
Evaluating f'(5) by substituting x = 5 into the simplified expression for f'(x).

The final result, f'(5) = 660, represents the instantaneous rate of change of the function at x = 5. This problem highlights the importance of understanding and applying the fundamental rules of differentiation, such as the product rule and the power rule. These rules are essential tools in calculus for analyzing the behavior of functions and solving a wide range of problems in mathematics, physics, engineering, and other fields. The ability to find derivatives and evaluate them at specific points is a crucial skill for anyone working with mathematical models and wanting to understand how quantities change and interact with each other. This step-by-step solution provides a clear and concise guide to solving this type of problem, emphasizing the logical progression and the application of key concepts in calculus.

By following these steps and understanding the underlying principles, you can confidently tackle similar problems involving derivatives and their applications.