How To Find The Derivative Of F(x) = X²(x - 3)⁹ A Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into a classic calculus problem: finding the derivative of a function using the product rule and the chain rule. Don't worry, it's not as scary as it sounds! We'll break it down step by step so you can follow along easily. Our function is f(x) = x²(x - 3)⁹. Let's get started!

Understanding the Problem

Before we jump into the solution, it's crucial to understand what we're trying to achieve. We're tasked with finding the derivative of the function f(x) = x²(x - 3)⁹. In simpler terms, we want to find the instantaneous rate of change of this function at any given point. This is a fundamental concept in calculus and has applications in various fields, from physics to economics.

To find this derivative, we'll need to employ a couple of key calculus rules: the product rule and the chain rule. These rules are essential tools in our calculus toolbox, allowing us to differentiate complex functions that are made up of multiple parts. The product rule helps us when we have two functions multiplied together, like our x² and (x - 3)⁹. The chain rule, on the other hand, comes into play when we have a function inside another function, such as the (x - 3) raised to the power of 9.

Let's break down these rules a bit further. The product rule states that the derivative of the product of two functions, say u(x) and v(x), is given by (uv)' = u'v + uv'. In our case, we can think of x² as u(x) and (x - 3)⁹ as v(x). This means we'll need to find the derivatives of both x² and (x - 3)⁹ separately and then plug them into the product rule formula.

Now, let's consider the chain rule. This rule is used when we have a composite function, which is a function within a function. The chain rule states that the derivative of a composite function f(g(x)) is given by f'(g(x)) * g'(x). In our case, (x - 3)⁹ is a composite function because we have the function (x - 3) raised to the power of 9. To apply the chain rule, we'll need to differentiate the outer function (something to the power of 9) and then multiply it by the derivative of the inner function (x - 3).

So, you see, finding the derivative of f(x) = x²(x - 3)⁹ isn't just about applying a single formula. It's about understanding the structure of the function and recognizing which rules to apply and when. Once we've mastered these rules, we can tackle a wide range of differentiation problems with confidence. We must be meticulous in our calculations to ensure we arrive at the correct derivative. A small error in applying either the product rule or the chain rule can lead to a completely different result.

With a solid grasp of the problem and the tools we need, let's move on to the solution. We'll take it step by step, showing each calculation clearly so you can see exactly how we arrive at the final answer. Remember, the key to mastering calculus is practice, so don't be afraid to try similar problems on your own. The more you practice, the more comfortable you'll become with these concepts and techniques.

Applying the Product Rule

Alright, let's dive into the first step: applying the product rule. As we discussed earlier, the product rule helps us find the derivative of a function that is the product of two other functions. In our case, f(x) = x²(x - 3)⁹, we can identify two functions: u(x) = x² and v(x) = (x - 3)⁹. The product rule states that f'(x) = u'(x)v(x) + u(x)v'(x).

So, the first thing we need to do is find the derivatives of u(x) and v(x) separately. Let's start with u(x) = x². This is a simple power rule application. The power rule states that if we have a function of the form xⁿ, its derivative is nxⁿ⁻¹. Applying this to u(x) = x², we get u'(x) = 2x²⁻¹ = 2x. Easy peasy!

Now, let's tackle v(x) = (x - 3)⁹. This one is a bit more interesting because it requires us to use the chain rule in addition to the power rule. Remember, the chain rule is used when we have a composite function, a function within a function. In this case, we have the function (x - 3) raised to the power of 9. The chain rule states that if we have a function of the form f(g(x)), its derivative is f'(g(x)) * g'(x).

Applying the chain rule to v(x) = (x - 3)⁹, we first differentiate the outer function, which is something to the power of 9. Using the power rule, we get 9(x - 3)⁸. Then, we need to multiply this by the derivative of the inner function, which is (x - 3). The derivative of (x - 3) with respect to x is simply 1. So, v'(x) = 9(x - 3)⁸ * 1 = 9(x - 3)⁸.

Great! We've found both u'(x) and v'(x). Now we have all the pieces we need to plug into the product rule formula. We have u'(x) = 2x, v(x) = (x - 3)⁹, u(x) = x², and v'(x) = 9(x - 3)⁸. Plugging these into the product rule formula f'(x) = u'(x)v(x) + u(x)v'(x), we get:

f'(x) = (2x)(x - 3)⁹ + (x²)(9(x - 3)⁸)

This is the derivative, but it's not in its simplest form yet. We can simplify it further by factoring out common terms. This will make the expression cleaner and easier to work with. Factoring out common terms is an essential skill in calculus and algebra, and it often helps us to see the underlying structure of an expression more clearly.

So, let's move on to the next step: simplifying the derivative. We'll look for common factors in the two terms of our derivative and pull them out. This will not only make the expression look nicer, but it can also help us in further calculations, such as finding critical points or analyzing the function's behavior.

Simplifying the Derivative

Okay, guys, we've got our derivative: f'(x) = (2x)(x - 3)⁹ + (x²)(9(x - 3)⁸). It looks a bit clunky, right? Let's simplify this expression to make it more manageable. The key here is to identify common factors in both terms and factor them out. This process not only makes the expression look cleaner but also reveals the underlying structure, which can be helpful for further analysis.

Looking at the two terms, (2x)(x - 3)⁹ and (x²)(9(x - 3)⁸), we can see that both terms have a factor of x and a factor of (x - 3). The lowest power of x present in both terms is x¹ (or simply x), and the lowest power of (x - 3) is (x - 3)⁸. So, these are the factors we can pull out.

Let's factor out x(x - 3)⁸ from both terms. When we factor x(x - 3)⁸ from (2x)(x - 3)⁹, we're left with 2(x - 3). And when we factor x(x - 3)⁸ from (x²)(9(x - 3)⁸), we're left with 9x. So, our factored expression looks like this:

f'(x) = x(x - 3)⁸ [2(x - 3) + 9x]

See how much cleaner that looks already? We've taken a somewhat complicated expression and broken it down into a product of simpler terms. This makes it easier to work with and understand.

But we're not quite done yet! We can simplify the expression inside the brackets further. Let's distribute the 2 in 2(x - 3) and then combine like terms:

2(x - 3) = 2x - 6

So, our expression inside the brackets becomes:

2x - 6 + 9x

Now, we can combine the 2x and 9x terms:

2x - 6 + 9x = 11x - 6

So, the expression inside the brackets simplifies to 11x - 6. Now, let's put it all together. Our simplified derivative is:

f'(x) = x(x - 3)⁸ (11x - 6)

This is the final simplified form of the derivative. We've successfully applied the product rule, the chain rule, and factoring techniques to arrive at this answer. Notice how much simpler this expression is compared to the one we had after applying the product rule initially. This simplification makes it easier to analyze the function's behavior, such as finding critical points or intervals of increase and decrease.

Simplifying the derivative is an essential step in many calculus problems. It's not just about making the expression look nicer; it's about making it more useful for further analysis. A simplified derivative is easier to work with when finding critical points, determining concavity, or solving optimization problems. So, mastering these simplification techniques is a crucial part of your calculus journey.

Final Answer

Alright, we've reached the end of our journey! We started with the function f(x) = x²(x - 3)⁹, and after carefully applying the product rule, the chain rule, and some algebraic simplification, we've arrived at the derivative:

f'(x) = x(x - 3)⁸ (11x - 6)

This is our final answer, and it represents the instantaneous rate of change of the function f(x) at any given point x. It's a powerful result that we can use to analyze the behavior of the function, find its critical points, and much more.

Let's take a moment to appreciate the steps we took to get here. We started by recognizing that the function was a product of two other functions, which led us to apply the product rule. Then, we encountered a composite function within the product, which required us to use the chain rule. Finally, we simplified the resulting expression by factoring out common terms and combining like terms.

Each of these steps is a fundamental technique in calculus, and mastering them is essential for success in this field. The product rule and the chain rule are like the bread and butter of differentiation, and the ability to simplify expressions is crucial for making sense of the results.

Now, you might be wondering, what can we do with this derivative? Well, the possibilities are vast! We can use it to find the critical points of the function, which are the points where the derivative is either zero or undefined. These points are important because they often correspond to local maxima, local minima, or saddle points of the function.

We can also use the derivative to determine the intervals where the function is increasing or decreasing. If the derivative is positive on an interval, the function is increasing on that interval. If the derivative is negative, the function is decreasing. This information can help us sketch the graph of the function and understand its overall behavior.

Furthermore, we can take the derivative of the derivative (the second derivative) to analyze the concavity of the function. The second derivative tells us whether the function is concave up or concave down, which gives us even more insight into its shape.

So, finding the derivative is just the first step in a much larger process of analyzing and understanding a function. It's a fundamental tool that opens up a wide range of possibilities.

In conclusion, we've successfully found the derivative of f(x) = x²(x - 3)⁹ using the product rule, the chain rule, and simplification techniques. We've also discussed the importance of these techniques and how the derivative can be used to analyze the behavior of the function. Keep practicing, and you'll become a calculus master in no time!