Derivative Of H(x) = 3 / (x³ - 2x)²: Step-by-Step Guide
Hey guys! Today, we're diving into the fascinating world of calculus to tackle a common problem: finding the derivative of a function. Specifically, we're going to break down how to find the derivative of the function h(x) = 3 / (x³ - 2x)². This might look a bit intimidating at first, but don't worry! We'll take it step by step, using the power of calculus rules to make it super clear. Whether you're a student grappling with calculus concepts or just someone curious about how derivatives work, this guide is for you. So, let's jump right in and unravel this derivative together!
Understanding Derivatives
Before we dive into the specifics of our function, it's crucial to have a solid grasp of what derivatives actually are. In simple terms, a derivative measures the instantaneous rate of change of a function. Think of it as the slope of a curve at a specific point. If you've ever wondered how quickly something is changing – whether it's the speed of a car, the growth of a population, or the slope of a mathematical function – derivatives are your key tool.
The Power Rule and Chain Rule
To master derivatives, you've got to know the fundamental rules. Two of the most important ones we'll be using today are the power rule and the chain rule. Let's break them down:
- The Power Rule: This rule is your best friend when dealing with terms like x raised to a power. It states that if you have a function f(x) = xⁿ, its derivative f'(x) is n * x^(n-1). Essentially, you bring the exponent down and multiply it by x, then reduce the exponent by 1. This rule is a cornerstone of differentiation and will come in handy frequently.
- The Chain Rule: Now, things get interesting when we have a function inside another function, like in our example today. That's where the chain rule steps in. The chain rule states that if you have a composite function f(g(x)), its derivative is f'(g(x)) * g'(x). In simpler terms, you take the derivative of the outer function, keeping the inner function as is, and then multiply by the derivative of the inner function. This rule is essential for handling more complex functions and is a bit like peeling an onion – you deal with the outer layers first and then work your way inwards.
Why Are Derivatives Important?
You might be wondering, "Okay, derivatives sound cool, but why should I care?" Well, derivatives have a vast range of applications across various fields. In physics, they help us understand motion and forces. In economics, they're used to optimize profits and model market behavior. In computer science, they play a role in machine learning algorithms. From engineering to finance, derivatives are a fundamental tool for anyone dealing with change and optimization. Understanding derivatives opens doors to solving real-world problems and making informed decisions, making them a crucial part of any STEM education and beyond. So, mastering these concepts isn't just about passing a calculus test; it's about gaining a powerful problem-solving skill that you can apply in countless situations.
Setting Up the Problem
Alright, let's get down to business and tackle our specific problem: finding the derivative of h(x) = 3 / (x³ - 2x)². The first step in solving any calculus problem is to make sure the function is in a form that's easy to work with. Our function looks a bit complex as it is, so we're going to rewrite it to make it more manageable for differentiation.
Rewriting the Function
The key to simplifying our function lies in recognizing that dividing by a term is the same as multiplying by its inverse. In other words, we can rewrite 3 / (x³ - 2x)² as 3 * (x³ - 2x)^(-2). This transformation might seem small, but it's a game-changer because it brings the function into a form where we can easily apply the chain rule. By expressing the denominator as a negative exponent, we've set ourselves up for a smoother differentiation process. This is a common trick in calculus – always look for ways to rewrite functions to make them more amenable to the rules we have.
Identifying the Outer and Inner Functions
Now that we've rewritten our function as h(x) = 3 * (x³ - 2x)^(-2), the next step is to identify the outer and inner functions. This is crucial for applying the chain rule effectively. Think of it like peeling an onion: we need to identify the layers we're going to peel.
- Outer Function: The outer function is the main operation being performed. In our case, it's the power function – something raised to the power of -2. We can think of the outer function as f(u) = 3 * u^(-2), where u is some expression.
- Inner Function: The inner function is what's inside the parentheses, the expression being raised to the power. Here, the inner function is g(x) = x³ - 2x. This is the function that's being plugged into the outer function.
By clearly identifying these components, we've set the stage for a successful application of the chain rule. We know we'll need to differentiate both the outer and inner functions, and then combine them in the correct way. This step-by-step approach is what makes calculus problems less daunting and more approachable. So, with our functions neatly identified, let's move on to the next step: taking the derivatives!
Applying the Chain Rule
Okay, guys, we've laid the groundwork, and now it's time for the main event: applying the chain rule to find the derivative of our function, h(x) = 3 * (x³ - 2x)^(-2). Remember, the chain rule is our go-to tool when dealing with composite functions – functions nested inside other functions.
Differentiating the Outer Function
First up, let's tackle the outer function, which we identified as f(u) = 3 * u^(-2). To find its derivative, f'(u), we'll use the power rule. The power rule states that if f(u) = a * uⁿ, then f'(u) = n * a * u^(n-1). Applying this to our outer function:
- f'(u) = -2 * 3 * u^(-2-1)
- f'(u) = -6 * u^(-3)
So, the derivative of our outer function is -6u^(-3). Don't forget, u represents our inner function, x³ - 2x, but we'll plug that back in later. For now, we've successfully differentiated the outer layer of our function.
Differentiating the Inner Function
Next, we need to find the derivative of the inner function, g(x) = x³ - 2x. This is a straightforward application of the power rule and the constant multiple rule. Let's break it down:
- The derivative of x³ is 3x² (using the power rule).
- The derivative of -2x is -2 (using the power rule and the constant multiple rule).
So, the derivative of our inner function, g'(x), is 3x² - 2. We've now successfully differentiated both the outer and inner functions, which means we're ready to bring it all together using the chain rule.
Combining the Derivatives
Now for the magic! The chain rule tells us that the derivative of h(x) = f(g(x)) is h'(x) = f'(g(x)) * g'(x). We've already found f'(u) and g'(x), so let's plug everything in:
- h'(x) = -6 * (x³ - 2x)^(-3) * (3x² - 2)
We've now found the derivative, but it's always good practice to simplify our result. Let's clean this up a bit.
Simplifying the Result
We've successfully applied the chain rule and found the derivative of our function, but let's be honest, it looks a bit messy. Simplifying the result not only makes it cleaner and easier to read but also helps in understanding the function's behavior. So, let's roll up our sleeves and simplify h'(x) = -6 * (x³ - 2x)^(-3) * (3x² - 2).
Dealing with the Negative Exponent
The first thing we can tackle is the negative exponent. Remember that a^(-n) is the same as 1 / aⁿ. Applying this to our function, we can rewrite (x³ - 2x)^(-3) as 1 / (x³ - 2x)³. This gets rid of the negative exponent and moves the term to the denominator, which is a good step towards simplification.
Combining Terms
Now, let's rewrite our derivative with this change:
- h'(x) = -6 * (1 / (x³ - 2x)³) * (3x² - 2)
We can now combine the terms by multiplying the numerators together:
- h'(x) = -6 * (3x² - 2) / (x³ - 2x)³
Factoring and Further Simplification
To further simplify, let's distribute the -6 in the numerator:
- h'(x) = (-18x² + 12) / (x³ - 2x)³
We can also factor out a common factor from the numerator. Both -18 and 12 are divisible by -6, so let's factor that out:
- h'(x) = -6(3x² - 2) / (x³ - 2x)³
At this point, we might also consider factoring the denominator, but in this case, it doesn't lead to any further significant simplification. So, we'll leave it as it is. Our simplified derivative is now:
- h'(x) = (-18x² + 12) / (x³ - 2x)³
This simplified form is much cleaner and easier to work with. We've successfully navigated the complexities of the chain rule and arrived at a neat, presentable solution. Great job, guys!
Final Answer
Alright, we've reached the end of our journey! After carefully applying the chain rule and simplifying the result, we've arrived at the final answer for the derivative of h(x) = 3 / (x³ - 2x)². Let's recap the key steps we took:
- Rewrote the function: We started by rewriting h(x) as 3 * (x³ - 2x)^(-2) to make it easier to differentiate.
- Identified outer and inner functions: We recognized the outer function as f(u) = 3 * u^(-2) and the inner function as g(x) = x³ - 2x.
- Applied the chain rule: We differentiated both the outer and inner functions and then combined them using the chain rule formula.
- Simplified the result: We cleaned up the expression by dealing with the negative exponent and combining terms.
The Derivative
So, after all that hard work, the final derivative of h(x) is:
h'(x) = (-18x² + 12) / (x³ - 2x)³
This is our final answer! We've successfully found the derivative of a complex function using the chain rule and simplification techniques. Understanding how to tackle these kinds of problems is a huge win for anyone studying calculus.
What's Next?
Now that we've nailed this derivative, you might be wondering, "What can I do with this?" Well, derivatives have numerous applications. You can use this derivative to find critical points, determine where the function is increasing or decreasing, and even sketch the graph of the function. The possibilities are vast! Keep practicing, and you'll become a derivative-solving pro in no time. Remember, calculus is all about understanding change, and derivatives are your key to unlocking that understanding. Great job today, guys! Keep up the awesome work, and happy calculating!