Derivatives Of F(x) = -4x² Ln(x) Calculation And Evaluation

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In the realm of calculus, understanding derivatives is crucial for analyzing the behavior of functions. Derivatives provide insights into a function's rate of change, allowing us to determine where it is increasing or decreasing, and to identify its critical points. This article delves into the intricacies of finding the derivative of a specific function, f(x) = -4x² ln(x), and evaluating it at a particular point. This exploration will not only enhance your understanding of derivative calculation but also showcase the practical application of calculus in analyzing logarithmic functions. The focus of our exploration is not merely on arriving at an answer but on comprehending the underlying principles and techniques that empower us to tackle similar problems with confidence. We embark on this journey with the goal of demystifying the process, breaking down each step, and providing a comprehensive understanding that extends beyond the immediate problem. Calculus, at its heart, is the study of change, and derivatives are the tools we use to quantify and analyze this change. By mastering the techniques of differentiation, we unlock the ability to model and understand real-world phenomena that involve rates of change, from the velocity of a moving object to the growth rate of a population. In the context of our function, f(x) = -4x² ln(x), the derivative will tell us how the function's value changes as x changes. This information can be invaluable in various applications, such as optimization problems where we seek to find the maximum or minimum value of the function. Our exploration begins with a careful examination of the function itself, identifying its components and the rules of differentiation that will be relevant to our task. We will then proceed step-by-step through the differentiation process, highlighting the application of the product rule and the chain rule, which are essential tools in the calculus toolkit. Finally, we will evaluate the derivative at a specific point, showcasing how the abstract concept of a derivative translates into a concrete numerical value that provides meaningful information about the function's behavior at that point. By the end of this journey, you will have not only a solution to the problem at hand but also a deeper appreciation for the power and elegance of calculus.

Finding the Derivative f'(x)

The core challenge lies in finding the derivative, f'(x), of the function f(x) = -4x² ln(x). To accomplish this, we must employ the product rule of differentiation. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. In mathematical notation, if we have two functions u(x) and v(x), then the derivative of their product is given by: (u(x)v(x))' = u'(x)v(x) + u(x)v'(x). Applying this rule is crucial for correctly differentiating functions that are formed by the product of other functions, such as the one we are dealing with here. In our case, we can identify u(x) = -4x² and v(x) = ln(x). Therefore, to find f'(x), we need to find the derivatives of u(x) and v(x) separately and then apply the product rule. First, let's find the derivative of u(x) = -4x². Using the power rule, which states that the derivative of x^n is nx^(n-1), we get u'(x) = -8x. This is a straightforward application of the power rule, where we multiply the coefficient -4 by the exponent 2 and then reduce the exponent by 1. Next, let's find the derivative of v(x) = ln(x). The derivative of the natural logarithm function, ln(x), is a fundamental result in calculus and is given by v'(x) = 1/x. This is a standard derivative that is essential to memorize for calculus problems involving logarithmic functions. Now that we have u'(x) and v'(x), we can apply the product rule to find f'(x). Substituting the expressions we found earlier, we get: f'(x) = u'(x)v(x) + u(x)v'(x) = (-8x)(ln(x)) + (-4x²)(1/x). This expression represents the derivative of f(x) and captures the rate of change of the function at any given point x. To simplify this expression, we can perform some algebraic manipulations. Specifically, we can simplify the second term by canceling out one factor of x in the numerator and denominator. This gives us: f'(x) = -8x ln(x) - 4x. This simplified form of the derivative is easier to work with and will be particularly useful when we need to evaluate the derivative at a specific point. In summary, the process of finding f'(x) involved identifying the product of two functions, applying the product rule, finding the derivatives of the individual functions, substituting these derivatives into the product rule formula, and then simplifying the resulting expression. This is a common pattern in calculus problems involving products of functions, and mastering this technique is essential for success in calculus.

Step-by-step Calculation of f'(x)

Let's break down the calculation of f'(x) step-by-step for clarity. We start with the function f(x) = -4x² ln(x). As we established earlier, this function is a product of two simpler functions: u(x) = -4x² and v(x) = ln(x). The derivative of a product requires the application of the product rule, which we will use meticulously to ensure accuracy. Our first step is to find the derivatives of the individual functions, u(x) and v(x). For u(x) = -4x², we apply the power rule of differentiation. The power rule states that if we have a function of the form x^n, its derivative is nx^(n-1). Applying this rule to u(x), we multiply the coefficient -4 by the exponent 2, and then reduce the exponent by 1. This gives us u'(x) = -8x. This is a straightforward application of the power rule, and it is essential to understand this rule thoroughly as it is frequently used in calculus problems. Next, we find the derivative of v(x) = ln(x). The derivative of the natural logarithm function is a fundamental result in calculus. The derivative of ln(x) is simply 1/x. Thus, v'(x) = 1/x. This derivative is a standard result that is worth memorizing as it appears frequently in calculus problems involving logarithmic functions. Now that we have found u'(x) and v'(x), we can apply the product rule to find f'(x). The product rule states that (u(x)v(x))' = u'(x)v(x) + u(x)v'(x). Substituting the expressions we found earlier, we get: f'(x) = (-8x)(ln(x)) + (-4x²)(1/x). This is the direct application of the product rule, and it is crucial to ensure that we substitute the correct expressions into the formula. The next step is to simplify the expression we obtained. We can simplify the second term by canceling out one factor of x in the numerator and denominator. This gives us: f'(x) = -8x ln(x) - 4x. This simplified form of the derivative is easier to work with and provides a more concise representation of the rate of change of the function. In summary, the step-by-step calculation of f'(x) involved: 1. Identifying the function as a product of two simpler functions. 2. Applying the product rule of differentiation. 3. Finding the derivatives of the individual functions using the power rule and the derivative of the natural logarithm function. 4. Substituting these derivatives into the product rule formula. 5. Simplifying the resulting expression. By breaking down the process into these steps, we can ensure that we understand each stage of the calculation and minimize the risk of errors. This methodical approach is essential for success in calculus.

Evaluating f'(e³)

Now, let's proceed to the evaluation of f'(e³). This step involves substituting x = e³ into the derivative we found earlier, f'(x) = -8x ln(x) - 4x. This process allows us to determine the rate of change of the function f(x) at the specific point x = e³. The value of the derivative at a particular point provides valuable information about the function's behavior, such as whether it is increasing or decreasing at that point and how rapidly it is changing. Substituting x = e³ into the expression for f'(x), we get: f'(e³) = -8(e³) ln(e³) - 4(e³). This is a direct substitution, and it is crucial to ensure that we replace every instance of x with e³. The next step is to simplify this expression. We can simplify the logarithmic term, ln(e³), using the property of logarithms that ln(a^b) = b ln(a). In this case, ln(e³) = 3 ln(e). Since ln(e) = 1, we have ln(e³) = 3. This simplification is essential for obtaining a manageable expression for f'(e³). Substituting this simplified value back into the expression, we get: f'(e³) = -8(e³)(3) - 4(e³) = -24e³ - 4e³. Now, we can combine the terms since they both involve e³. This gives us: f'(e³) = -28e³. This is the exact value of the derivative at x = e³. However, the problem asks for the answer rounded to three decimal places. To obtain this, we need to approximate the value of e³. The value of e (Euler's number) is approximately 2.71828. Therefore, e³ is approximately 2.71828³. Calculating this value, we get e³ ≈ 20.0855. Substituting this approximation into the expression for f'(e³), we get: f'(e³) ≈ -28(20.0855) ≈ -562.394. Rounding this value to three decimal places, we get: f'(e³) ≈ -562.394. This is the final answer, rounded to the required precision. In summary, the process of evaluating f'(e³) involved: 1. Substituting x = e³ into the expression for f'(x). 2. Simplifying the logarithmic term using the properties of logarithms. 3. Approximating the value of e³. 4. Substituting the approximation into the expression and calculating the result. 5. Rounding the result to three decimal places. This process demonstrates how we can use the derivative to find the rate of change of a function at a specific point and how we can approximate the value of the derivative to a desired level of precision.

Numerical Approximation of f'(e³)

To further clarify the evaluation, let's focus on the numerical approximation of f'(e³). As we derived earlier, f'(e³) = -28e³. The key to obtaining a numerical approximation lies in understanding the value of e, Euler's number, which is approximately 2.71828. This constant plays a fundamental role in calculus and appears in many mathematical and scientific contexts. To evaluate e³, we need to raise this approximate value of e to the power of 3. Using a calculator or a computer, we find that e³ ≈ 20.085536923187668. This is a more precise value of e³ than we used in the previous section, and using this value will give us a more accurate approximation of f'(e³). Now, we substitute this value into our expression for f'(e³): f'(e³) = -28e³ ≈ -28(20.085536923187668). Performing this multiplication, we get: f'(e³) ≈ -562.3950338492547. The problem asks for the answer rounded to three decimal places. To round this number to three decimal places, we look at the fourth decimal place. If it is 5 or greater, we round up the third decimal place. If it is less than 5, we leave the third decimal place as it is. In this case, the fourth decimal place is 0, which is less than 5, so we leave the third decimal place as it is. Therefore, rounding to three decimal places, we get: f'(e³) ≈ -562.395. This is the numerical approximation of f'(e³) rounded to three decimal places. It is important to note that the accuracy of our approximation depends on the accuracy of the value we use for e³. The more decimal places we use for e, the more accurate our approximation will be. In practice, calculators and computers typically use a large number of decimal places for e, so the approximation we obtain is usually very accurate. In summary, the numerical approximation of f'(e³) involved: 1. Recalling the approximate value of e (Euler's number). 2. Calculating e³ using a calculator or computer. 3. Substituting the value of e³ into the expression for f'(e³). 4. Performing the multiplication. 5. Rounding the result to three decimal places. This process demonstrates how we can use numerical approximations to evaluate expressions in calculus and how we can obtain results to a desired level of precision. The ability to perform numerical approximations is an essential skill in calculus and is widely used in various applications.

Conclusion

In conclusion, we have successfully navigated the process of finding the derivative of f(x) = -4x² ln(x) and evaluating it at x = e³. This journey has underscored the importance of the product rule in differentiation, the fundamental derivatives of logarithmic functions, and the practical application of these concepts in evaluating derivatives at specific points. The step-by-step approach we employed, from identifying the component functions to simplifying the final expression, highlights the systematic nature of calculus and the power of breaking down complex problems into manageable steps. The final answer, f'(e³) ≈ -562.395, not only provides a numerical value but also carries significant information about the function's behavior at x = e³. It tells us the instantaneous rate of change of the function at this point, which can be crucial in various applications such as optimization problems or modeling physical phenomena. Moreover, the process of numerical approximation emphasizes the practical side of calculus. While exact solutions are often desirable, they are not always attainable, and numerical methods provide a powerful tool for obtaining approximate solutions to a desired level of precision. This is particularly relevant in real-world applications where data may be imprecise or where computational limitations exist. The skills and concepts explored in this article are not isolated to this specific problem. They form the bedrock of calculus and are applicable to a wide range of problems involving derivatives. Mastering these techniques will empower you to tackle more complex problems and to apply calculus in diverse fields. The journey through this problem has also highlighted the beauty and elegance of calculus. The way that the derivative captures the essence of change, the way that seemingly complex functions can be differentiated using systematic rules, and the way that numerical methods can provide approximate solutions all contribute to the rich tapestry of calculus. As you continue your exploration of calculus, remember that the key to success lies in understanding the underlying principles, practicing the techniques, and appreciating the power and beauty of this fundamental branch of mathematics.