Calculating H'(-1): A Step-by-Step Derivative Guide

Hey math enthusiasts! Today, we're diving into the world of derivatives, specifically focusing on how to calculate h'(-1) for the function h(x) = x³(x²+1)⁴. Don't worry if you're feeling a bit rusty on your calculus skills; we'll break down each step, making it super easy to follow. This guide will walk you through everything, from the product rule to simplifying the final answer. So, grab your pencils, and let's get started! Understanding derivatives is crucial in calculus, as they represent the instantaneous rate of change of a function. In simpler terms, the derivative tells us how much the output of a function changes with respect to an infinitesimally small change in its input. We'll be using a combination of the product rule and the chain rule to solve this problem. This is a common type of problem, so mastering the steps here will set you up for success in your calculus journey. The goal is not just to find the answer but to truly grasp the underlying principles and the reasoning behind each step. Let's get into it, guys!

Understanding the Problem: The Function and Our Goal

First things first, let's clarify what we're dealing with. We have the function h(x) = x³(x²+1)⁴. This is a product of two functions: x³ and (x²+1)⁴. Our goal is to find the value of the derivative of h(x), which we denote as h'(x), at the point x = -1. This means we need to calculate the derivative of the function, and then plug in -1 into the resulting equation. Remember, the derivative at a specific point gives us the slope of the tangent line to the function at that point. It’s a fundamental concept in calculus, helping us understand the behavior of functions. The product rule and chain rule will be our primary tools. So, what's the product rule? It's a formula that helps us find the derivative of a product of two functions. The chain rule, on the other hand, is used when we have a function within another function (a composite function), like we do with (x²+1)⁴. Before we get to the calculations, let’s refresh on the basic derivative rules. You'll need to know how to find the derivative of power functions (like x³), and the derivative of a constant, and how to work with composite functions. By knowing these rules, you are well on your way to conquering this problem and similar derivative problems.

Now, let's move on to the next part, where we begin to compute the derivative.

Applying the Product Rule: The First Step

Alright, let's get our hands dirty and start calculating the derivative of h(x). Since our function is a product of two other functions, we'll use the product rule. The product rule states that if we have a function f(x) = u(x) * v(x), then its derivative f'(x) = u'(x) * v(x) + u(x) * v'(x). In our case, u(x) = x³ and v(x) = (x²+1)⁴. Let's find the derivatives of u(x) and v(x) separately. First, u'(x), the derivative of x³, is simply 3x². Easy peasy, right? Now, things get a little more interesting with v(x) = (x²+1)⁴. Here, we'll need to use the chain rule, since we have a function raised to a power. The chain rule says that if we have a composite function f(g(x)), then its derivative f'(g(x)) * g'(x). This is why it's so important to know the basics of the product and chain rule because we will need to apply both of them to the problem. The chain rule helps us break down the derivative of composite functions, making them easier to compute. Applying the product rule first sets the stage for breaking down the problem into manageable steps. By identifying u(x) and v(x) and their respective derivatives, we establish a framework for applying the product rule correctly. Each step builds on the previous one, ensuring that our method is not only accurate but also easy to follow. Now, let's move on to the next part of the calculation.

So, are you ready to solve the next step?

Applying the Chain Rule: Diving Deeper into the Derivative

Now, let's tackle the derivative of v(x) = (x²+1)⁴ using the chain rule. We're essentially dealing with an outer function (something raised to the power of 4) and an inner function (x²+1). According to the chain rule, the derivative of (x²+1)⁴ is 4(x²+1)³ * 2x. Why? Well, the derivative of the outer function is 4 times the inner function to the power of 3, multiplied by the derivative of the inner function (which is 2x). We use the chain rule when we have one function inside another. It's a fundamental technique to find the derivative of composite functions. The correct use of the chain rule is very important in this step, because this can easily be done incorrectly. It allows us to decompose complex functions into manageable parts and find the derivative of those parts separately and combine them appropriately. Keep in mind to multiply the derivative of the outer function with the derivative of the inner function. It's this multiplication that captures how the change in the inner function impacts the overall function's change. Now that we have u'(x) = 3x² and v'(x) = 4(x²+1)³ * 2x = 8x(x²+1)³, we can put everything back together using the product rule formula: h'(x) = u'(x) * v(x) + u(x) * v'(x). This gives us h'(x) = 3x²(x²+1)⁴ + x³ * 8x(x²+1)³. Great job guys!

Let’s proceed to simplify the above equations.

Simplifying the Derivative: Combining and Factoring

We now have the derivative, h'(x) = 3x²(x²+1)⁴ + x³ * 8x(x²+1)³. Before we plug in x = -1, let's simplify this expression to make our lives easier. First, we can factor out common terms. Notice that both terms have x² and (x²+1)³. So, let's factor those out. We can rewrite h'(x) as x²(x²+1)³ [3(x²+1) + 8x²]. This simplification is helpful for the following steps, it makes our calculations much easier. Factoring allows us to reduce the complexity of the expression, making it easier to evaluate at x = -1. We are able to combine like terms and reduce the degree of the polynomial if possible, making the evaluation process faster. Factoring helps us see the structure of the derivative more clearly. This structure will help us in determining the derivative at a specific point. Factoring also reduces the likelihood of calculation errors. It’s a really essential skill in calculus, streamlining complex expressions into more manageable forms. Make sure you don’t skip this step. By now, the expression inside the brackets is equal to 3x² + 3 + 8x² = 11x² + 3. Thus, our simplified derivative is h'(x) = x²(x²+1)³(11x² + 3). This is the simplified version of the derivative function that we are going to use. Now, let's finally calculate h'(-1)!

Let’s see how this is calculated in the next step.

Evaluating h'(-1): The Final Calculation

Now comes the moment of truth: evaluating h'(-1). We have our simplified derivative, h'(x) = x²(x²+1)³(11x² + 3). Let's substitute x = -1 into this expression. We get h'(-1) = (-1)²((-1)²+1)³(11(-1)² + 3). Remember, when we substitute a negative number, we use parentheses to make sure our calculations are correct. This avoids any common arithmetic mistakes. Now, let's simplify this further. (-1)² = 1, ((-1)²+1) = 2, and (11(-1)² + 3) = 14. So, our expression becomes h'(-1) = 1 * 2³ * 14. Now, calculate 2³ is 8, and 1 * 8 * 14 = 112. Therefore, h'(-1) = 112. We've done it! It’s a great achievement to successfully complete the problem. By breaking down the derivative into manageable steps, we've made the process much less daunting. We’ve not only found the correct answer but also deepened our understanding of derivatives. Keep practicing, and you'll become a pro at these types of problems. Also, if you get stuck, don’t hesitate to review the steps and examples! It's all about practicing and learning. That's it for today's lesson. See you next time, guys!

Summary of Steps

1. Identify the Function: We started with h(x) = x³(x²+1)⁴. Identify the product of two functions, u(x) = x³ and v(x) = (x²+1)⁴.
2. Apply the Product Rule: h'(x) = u'(x)v(x) + u(x)v'(x).
3. Find the Derivatives:
   • u'(x) = 3x²
   • v'(x) = 8x(x²+1)³ (using the chain rule)
4. Substitute and Simplify: h'(x) = 3x²(x²+1)⁴ + x³ * 8x(x²+1)³ → h'(x) = x²(x²+1)³(11x² + 3).
5. Evaluate at x = -1: h'(-1) = (-1)²((-1)²+1)³(11(-1)² + 3) = 112.

Conclusion

We've successfully calculated h'(-1) for the function h(x) = x³(x²+1)⁴. We used the product rule and the chain rule, then simplified the resulting expression and evaluated it at x = -1. The key takeaways are the product rule, the chain rule, and simplification techniques. Remember, practice makes perfect. If you are just starting out, you should consider working on basic functions and rules first. Keep in mind, the more problems you solve, the more confident you'll become in your calculus skills. Keep up the great work, and happy calculating!