Finding Derivatives Using Limits A Comprehensive Guide

Hey everyone! Today, we're diving into the exciting world of calculus, where we'll explore how to use limits to find the derivative of a function. Specifically, we'll be working with the function f(x) = 5x² - 4x. Don't worry if that looks intimidating – we'll break it down step by step, making it super easy to understand. We will also be evaluating f’(a) for the given values of a.

a. Finding the Derivative Function f'(x) Using Limits

So, what exactly is a derivative? In simple terms, the derivative of a function at a point tells us the instantaneous rate of change of the function at that point. Think of it like the speedometer in your car – it shows your speed at a specific moment in time, not your average speed over a whole trip. To find this instantaneous rate of change, we use the concept of limits. Guys, this is where things get interesting!

The formal definition of the derivative, using limits, is this:

f'(x) = lim (h->0) [f(x + h) - f(x)] / h

This formula might look a bit scary, but let's break it down:

• f'(x): This is the notation for the derivative of the function f(x).
• lim (h->0): This means we're taking the limit as h approaches 0. h represents a tiny change in x.
• f(x + h): This is the value of the function when we plug in x + h instead of x.
• f(x + h) - f(x): This represents the change in the function's value when x changes by h.
• [f(x + h) - f(x)] / h: This is the average rate of change of the function over the small interval h.

By taking the limit as h approaches 0, we're essentially squeezing this interval down to a single point, giving us the instantaneous rate of change, which is the derivative.

Okay, let's apply this to our function, f(x) = 5x² - 4x. Buckle up, guys, it's calculation time!

1. Find f(x + h):

   We replace x with (x + h) in the function:

   f(x + h) = 5(x + h)² - 4(x + h)

   Now, we expand and simplify:

   f(x + h) = 5(x² + 2xh + h²) - 4x - 4h

   f(x + h) = 5x² + 10xh + 5h² - 4x - 4h

2. Calculate f(x + h) - f(x):

   We subtract the original function, f(x) = 5x² - 4x, from the expression we just found:

   f(x + h) - f(x) = (5x² + 10xh + 5h² - 4x - 4h) - (5x² - 4x)

   Notice how some terms cancel out? This is a good sign!

   f(x + h) - f(x) = 10xh + 5h² - 4h

3. Divide by h:

   We divide the result by h:

   [f(x + h) - f(x)] / h = (10xh + 5h² - 4h) / h

   We can factor out an h from the numerator:

   [f(x + h) - f(x)] / h = h(10x + 5h - 4) / h

   Now, we can cancel the h terms:

   [f(x + h) - f(x)] / h = 10x + 5h - 4

4. Take the limit as h approaches 0:

   This is the final step! We take the limit of the expression as h gets closer and closer to 0:

   f'(x) = lim (h->0) (10x + 5h - 4)

   As h approaches 0, the term 5h also approaches 0. So, we're left with:

   f'(x) = 10x - 4

   And there you have it! We've found the derivative function of f(x) = 5x² - 4x using the limit definition. The derivative, f'(x), is 10x - 4. This tells us the slope of the tangent line to the graph of f(x) at any point x. Awesome, right?

b. Evaluating f'(a) for a = -4 and a = 2

Now that we have the derivative function, f'(x) = 10x - 4, we can easily find the instantaneous rate of change at specific points. The problem asks us to evaluate f'(a) for a = -4 and a = 2. This simply means we need to plug in these values for x in our derivative function.

Evaluating f'(-4)

Let's start with a = -4. We substitute x with -4 in f'(x) = 10x - 4:

f'(-4) = 10(-4) - 4

f'(-4) = -40 - 4

f'(-4) = -44

So, the derivative of the function f(x) at x = -4 is -44. This means that at the point x = -4 on the graph of f(x), the tangent line has a slope of -44. It's a pretty steep slope, guys!

Evaluating f'(2)

Now, let's evaluate f'(2). We substitute x with 2 in f'(x) = 10x - 4:

f'(2) = 10(2) - 4

f'(2) = 20 - 4

f'(2) = 16

Therefore, the derivative of the function f(x) at x = 2 is 16. This means that at the point x = 2 on the graph of f(x), the tangent line has a slope of 16. This is a positive slope, indicating that the function is increasing at this point. See guys, it is not that difficult!

Significance of Derivatives

Derivatives are a fundamental concept in calculus with wide-ranging applications in various fields, including physics, engineering, economics, and computer science. They allow us to analyze the behavior of functions, such as finding their maximum and minimum values, determining their rate of change, and modeling real-world phenomena.

In physics, derivatives are used to describe velocity and acceleration. In engineering, they are used to optimize designs and analyze system stability. In economics, they are used to model supply and demand curves. In computer science, they are used in machine learning algorithms and computer graphics.

Understanding derivatives opens up a whole new world of possibilities for problem-solving and analysis. It's a powerful tool that can help us make sense of the world around us.

Key Takeaways

• The derivative of a function represents its instantaneous rate of change.
• We can find the derivative using the limit definition: f'(x) = lim (h->0) [f(x + h) - f(x)] / h.
• The derivative function, f'(x), gives us the slope of the tangent line to the graph of f(x) at any point x.
• Evaluating f'(a) gives us the instantaneous rate of change at a specific point x = a.
• Derivatives have numerous applications in various fields.

Conclusion

Guys, we've successfully navigated the process of finding the derivative of a function using limits and evaluating it at specific points. This is a crucial skill in calculus, and I hope this explanation has made it clear and easy to understand. Remember, practice makes perfect, so keep working on these problems, and you'll become a derivative pro in no time! Keep exploring the fascinating world of calculus, and you'll discover its incredible power and versatility. You've got this!