Derivative Of F(x) = X² * Ln(x) Step-by-Step With Justification

Hey everyone! Today, we're diving deep into the world of calculus to explore the derivative of a fascinating function: f(x) = x² * ln(x). This function combines a polynomial term (x²) with a logarithmic term (ln(x)), making its derivative a bit more interesting to calculate. But don't worry, we'll break it down step by step, so it's super easy to follow. Whether you're a student grappling with calculus, a math enthusiast, or just curious about how derivatives work, this guide is for you. We'll not only find the derivative but also understand the why behind each step, ensuring you grasp the underlying concepts. So, grab your pencils and notebooks, and let's embark on this mathematical journey together!

Understanding the Basics: Derivatives and the Product Rule

Before we jump into the specifics of f(x) = x² * ln(x), let's quickly revisit the core concept of a derivative. In simple terms, a derivative tells us the instantaneous rate of change of a function at a particular point. Think of it as the slope of the tangent line to the function's graph at that point. Derivatives are fundamental to calculus and have wide-ranging applications in physics, engineering, economics, and many other fields. They help us analyze how things change and make predictions based on those changes. To tackle our function, we'll need a crucial tool: the product rule. The product rule comes into play when we need to find the derivative of a function that is the product of two other functions. In our case, f(x) = x² * ln(x) is clearly a product of two functions: x² and ln(x). The product rule states that if we have a function h(x) = u(x) * v(x), then its derivative h'(x) is given by: h'(x) = u'(x) * v(x) + u(x) * v'(x). In plain English, the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. This might sound a bit complex, but it's a straightforward formula that becomes second nature with practice. So, let's keep this product rule in mind as our guiding principle as we move forward. Now that we've refreshed our understanding of derivatives and the product rule, we're well-equipped to tackle the derivative of f(x) = x² * ln(x). We'll identify our 'u(x)' and 'v(x)', find their individual derivatives, and then apply the product rule to piece everything together. Get ready to see how it all unfolds!

Step 1: Identifying u(x) and v(x)

The first step in applying the product rule to find the derivative of f(x) = x² * ln(x) is to correctly identify our two functions, u(x) and v(x). This is a crucial step because it sets the stage for the rest of our calculation. A wrong identification here can lead to an incorrect derivative, so let's make sure we get it right. Looking at our function, it's quite clear that it's composed of two distinct parts multiplied together: x² and ln(x). This makes our choice for u(x) and v(x) pretty straightforward. We can let u(x) = x² and v(x) = ln(x). It's as simple as that! We've successfully broken down our original function into two smaller, more manageable pieces. This is a common strategy in calculus – breaking down complex problems into simpler ones that we can solve individually. Now that we have our u(x) and v(x) clearly defined, the next step is to find their individual derivatives. This will involve applying some basic differentiation rules that you might already be familiar with. But if not, don't worry! We'll go through them step by step to ensure everyone's on the same page. Finding the derivatives of u(x) and v(x) is like gathering the ingredients for a recipe. Once we have these ingredients, we can use the product rule as our recipe to combine them in the correct way and arrive at the final derivative of f(x). So, let's move on to the next step and find those derivatives! Remember, the key here is to be methodical and take things one step at a time. Calculus can seem daunting at first, but by breaking it down into smaller, manageable steps, it becomes much more accessible and even enjoyable. Keep that in mind as we continue our journey towards finding the derivative of f(x) = x² * ln(x).

Step 2: Finding the Derivatives of u(x) and v(x)

Now that we've identified u(x) = x² and v(x) = ln(x), our next task is to find their respective derivatives, u'(x) and v'(x). This is where our knowledge of basic differentiation rules comes into play. Let's start with u(x) = x². To find its derivative, u'(x), we'll use the power rule. The power rule is a fundamental rule in calculus that states if we have a function of the form x^n, where n is any real number, then its derivative is given by n*x^(n-1). In other words, we bring the exponent down as a coefficient and then reduce the exponent by 1. Applying the power rule to u(x) = x², we get u'(x) = 2 * x^(2-1) = 2x. So, the derivative of x² is simply 2x. That wasn't so hard, was it? Now, let's move on to v(x) = ln(x). Finding the derivative of the natural logarithm function is another fundamental rule in calculus. The derivative of ln(x) is 1/x. So, v'(x) = 1/x. This is a rule that you'll likely encounter frequently in calculus, so it's a good one to memorize. We've now successfully found the derivatives of both u(x) and v(x). We have u'(x) = 2x and v'(x) = 1/x. These are the crucial ingredients we need to apply the product rule. Think of it like this: we've prepared all the individual components, and now we're ready to assemble them according to the recipe of the product rule. The next step is to plug these derivatives, along with our original functions u(x) and v(x), into the product rule formula. This will give us the derivative of f(x) = x² * ln(x). So, let's move on to the exciting part where we put everything together and see the final result! Remember, calculus is all about breaking down complex problems into simpler steps. We've done the hard work of identifying our functions and finding their derivatives, and now we're just one step away from the final answer.

Step 3: Applying the Product Rule

With u(x) = x², v(x) = ln(x), u'(x) = 2x, and v'(x) = 1/x in hand, we're finally ready to apply the product rule. As a quick reminder, the product rule states that if f(x) = u(x) * v(x), then f'(x) = u'(x) * v(x) + u(x) * v'(x). This is the formula that will guide us in combining the pieces we've already found. Now, let's plug in our values into the product rule formula. We have f'(x) = (2x) * (ln(x)) + (x²) * (1/x). This is the direct application of the product rule to our specific functions. We've simply substituted the expressions we found earlier into the formula. The next step is to simplify this expression to obtain a more concise and manageable form of the derivative. Simplification is a crucial part of calculus because it allows us to work with the derivative more easily and to better understand its behavior. Looking at our expression, we can see that there's some potential for simplification. In the second term, we have x² multiplied by 1/x. This can be simplified by canceling out a factor of x from the numerator and the denominator. So, let's do that in the next step. Remember, the goal of applying the product rule is not just to plug in the values but also to simplify the resulting expression. A simplified derivative is much easier to work with for further analysis, such as finding critical points or analyzing the function's concavity. So, let's move on to the simplification step and see how we can make our derivative look cleaner and more elegant. We're almost there! We've done the heavy lifting of finding the individual derivatives and applying the product rule. Now, it's just a matter of simplifying the expression to arrive at the final answer. Keep your eyes on the prize, and let's finish this strong!

Step 4: Simplifying the Expression

Now that we've applied the product rule and have f'(x) = (2x) * (ln(x)) + (x²) * (1/x), it's time to simplify this expression. Simplifying is not just about making the expression look neater; it also makes it easier to analyze and use in further calculations. The key to simplifying this expression lies in the second term: (x²) * (1/x). We have x² in the numerator and x in the denominator, which means we can cancel out one factor of x. When we cancel out a factor of x, we're left with just x in the second term. So, (x²) * (1/x) simplifies to x. Now, let's rewrite our derivative with this simplification: f'(x) = 2x * ln(x) + x. This looks much cleaner already! But we can go even further. Notice that both terms in our expression now have a common factor of x. We can factor out this x to simplify the expression even more. Factoring out x, we get: f'(x) = x * (2ln(x) + 1). And there we have it! We've successfully simplified the derivative of f(x) = x² * ln(x). Our final simplified derivative is f'(x) = x * (2ln(x) + 1). This is a much more concise and manageable expression than what we started with after applying the product rule. This simplified form makes it easier to analyze the behavior of the derivative, such as finding its zeros or determining where the function is increasing or decreasing. Remember, simplification is a crucial skill in calculus. It not only makes expressions look neater but also makes them easier to work with. By simplifying our derivative, we've made it much more useful for further analysis. So, pat yourselves on the back! We've successfully navigated the product rule and simplified the resulting expression. We've found the derivative of f(x) = x² * ln(x), and we've done it step by step, ensuring we understand each step along the way. Now, let's take a moment to appreciate what we've accomplished and think about how this derivative can be used in other contexts.

The Final Result: f'(x) = x * (2ln(x) + 1)

We've reached the end of our journey to find the derivative of f(x) = x² * ln(x), and the final result is: f'(x) = x * (2ln(x) + 1). This is the derivative we've been working towards, and it represents the instantaneous rate of change of the function f(x) at any given point x. It's a powerful result that can be used in a variety of applications. Let's take a moment to appreciate what this derivative tells us. It tells us how the function f(x) = x² * ln(x) is changing at any point. For example, we can use this derivative to find the slope of the tangent line to the graph of f(x) at any specific x-value. We can also use it to find the critical points of the function, which are the points where the derivative is either zero or undefined. These critical points can help us determine the maximum and minimum values of the function. Furthermore, we can use the derivative to analyze the concavity of the function, which tells us whether the graph of the function is curving upwards or downwards. This information can be used to sketch the graph of the function and to understand its overall behavior. The derivative f'(x) = x * (2ln(x) + 1) is not just a formula; it's a tool that provides us with valuable insights into the function f(x) = x² * ln(x). It's a testament to the power of calculus and its ability to help us understand the world around us. So, congratulations on making it to the end! You've successfully found the derivative of f(x) = x² * ln(x) and learned about its significance. This is a great accomplishment, and you should be proud of yourself. Remember, calculus is a journey of continuous learning and exploration. There's always more to discover, more to understand, and more to apply. So, keep practicing, keep exploring, and keep pushing your boundaries. The world of calculus is vast and fascinating, and it's waiting for you to uncover its secrets.

Practical Applications and Further Exploration

Now that we've successfully found the derivative of f(x) = x² * ln(x), let's take a moment to explore some practical applications and further avenues of exploration. Understanding how a derivative can be used in real-world scenarios is just as important as knowing how to calculate it. One of the most common applications of derivatives is in optimization problems. These are problems where we want to find the maximum or minimum value of a function, subject to certain constraints. For example, we might want to find the dimensions of a rectangular garden that maximize the area enclosed, given a fixed amount of fencing. To solve optimization problems, we often use the derivative to find the critical points of the function, which are the points where the derivative is zero or undefined. These critical points are potential locations of maximum or minimum values. Another important application of derivatives is in related rates problems. These are problems where we have two or more quantities that are changing with time, and we want to find the rate at which one quantity is changing in terms of the rates at which the other quantities are changing. For example, we might want to find the rate at which the volume of a balloon is increasing as we inflate it, given the rate at which the radius is increasing. To solve related rates problems, we use implicit differentiation and the chain rule to relate the derivatives of the different quantities. Beyond these specific applications, derivatives are also fundamental to many other areas of mathematics, science, and engineering. They are used in physics to describe velocity and acceleration, in economics to model marginal cost and revenue, and in computer science to develop machine learning algorithms. If you're interested in further exploring the world of calculus, there are many resources available to you. You can delve deeper into the theory of derivatives, explore different types of functions and their derivatives, and learn about more advanced techniques such as integration and differential equations. Calculus is a vast and fascinating subject, and there's always more to learn. So, keep practicing, keep exploring, and keep pushing your mathematical boundaries. The possibilities are endless!

Conclusion: Mastering Derivatives, One Step at a Time

Well, guys, we've reached the end of our step-by-step journey to find the derivative of f(x) = x² * ln(x)! We've not only calculated the derivative but also delved into the why behind each step. Remember, the final derivative is f'(x) = x * (2ln(x) + 1). This wasn't just about getting the right answer; it was about understanding the process and the underlying principles of calculus. We started by understanding the basics of derivatives and the crucial product rule. Then, we methodically broke down the problem into smaller, manageable steps: identifying u(x) and v(x), finding their individual derivatives, applying the product rule, and simplifying the resulting expression. This step-by-step approach is key to mastering calculus and any complex mathematical concept. By breaking down a problem into smaller parts, you make it less intimidating and easier to understand. We also explored some practical applications of derivatives, highlighting their importance in various fields. Derivatives aren't just abstract mathematical concepts; they're powerful tools that can help us solve real-world problems. If you found this guide helpful, that's awesome! Keep practicing and exploring different types of functions and their derivatives. The more you practice, the more comfortable and confident you'll become with calculus. And remember, if you ever get stuck, don't hesitate to break the problem down into smaller steps and revisit the fundamental concepts. Calculus can be challenging, but it's also incredibly rewarding. It's a journey of continuous learning and discovery, and it's one that's well worth taking. So, keep up the great work, and never stop exploring the fascinating world of mathematics! You've got this! And who knows, maybe you'll even start to enjoy finding derivatives as much as we do. Thanks for joining us on this mathematical adventure!