Finding The Derivative Of F(x) = 2 Arctan(3sin(4x)) A Calculus Exploration

In the realm of calculus, finding the derivative of a function is a fundamental operation. It allows us to understand the rate at which a function's output changes with respect to its input. This understanding is crucial in various fields, from physics and engineering to economics and computer science. In this article, we will delve into the process of finding the derivative of a specific function: f(x) = 2 arctan(3sin(4x)). This function involves trigonometric functions, inverse trigonometric functions, and the chain rule, making it an excellent example for illustrating the power and elegance of differential calculus. Our journey will begin with a brief review of the essential concepts and rules of differentiation, followed by a step-by-step application of these rules to the given function. We will meticulously break down each step, ensuring clarity and understanding. By the end of this exploration, you will not only grasp the solution to this particular problem but also gain a deeper appreciation for the techniques and principles involved in finding derivatives of complex functions.

Before we embark on the journey of finding the derivative of f(x) = 2 arctan(3sin(4x)), it's crucial to have a firm grasp of some fundamental calculus concepts. These concepts serve as the building blocks for our exploration, enabling us to navigate the intricacies of differentiation with confidence. Let's take a moment to refresh our understanding of these essential tools. First and foremost, we need to recall the chain rule. This powerful rule is our guiding light when differentiating composite functions – functions within functions. The chain rule states that the derivative of f(g(x)) with respect to x is f'(g(x)) * g'(x). In simpler terms, we differentiate the outer function, keeping the inner function intact, and then multiply by the derivative of the inner function. This rule is paramount in tackling functions like the one we're about to explore. Next, we must have a clear understanding of the derivatives of basic trigonometric functions, specifically sine and cosine. The derivative of sin(x) with respect to x is cos(x), and the derivative of cos(x) with respect to x is -sin(x). These derivatives are the foundation upon which we build the derivatives of more complex trigonometric expressions. In addition to trigonometric functions, we must also be familiar with the derivative of the inverse tangent function, also known as the arctangent function. The derivative of arctan(x) with respect to x is 1 / (1 + x^2). This derivative is a cornerstone in differentiating expressions involving arctangent, and it will play a pivotal role in our current problem. Lastly, let's not forget the constant multiple rule. This rule simply states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. In mathematical notation, if c is a constant, then the derivative of c * f(x) with respect to x is c * f'(x). This rule simplifies our calculations and allows us to focus on the core differentiation process. With these essential concepts firmly in mind, we are now well-equipped to tackle the challenge of finding the derivative of f(x) = 2 arctan(3sin(4x)).

Step-by-Step Differentiation of f(x) = 2 arctan(3sin(4x))

Now, let's dive into the process of finding the derivative of f(x) = 2 arctan(3sin(4x)) step-by-step. We will meticulously apply the rules and concepts we discussed earlier to unravel the derivative of this composite function. This methodical approach will not only lead us to the solution but also enhance our understanding of the differentiation process. Our journey begins by recognizing the structure of the function. We see that f(x) is a composite function, a function within a function. It consists of the arctangent function, the sine function, and a constant multiple. This immediately signals the importance of the chain rule in our approach. We will be applying the chain rule multiple times, peeling away the layers of the function one by one. The first layer we encounter is the constant multiple of 2. According to the constant multiple rule, we can simply carry this constant along throughout the differentiation process. So, the derivative of 2 * arctan(3sin(4x)) will be 2 times the derivative of arctan(3sin(4x)). This simplifies our task, allowing us to focus on the core composite function. Next, we tackle the arctangent function. Recall that the derivative of arctan(x) with respect to x is 1 / (1 + x^2). In our case, the argument of the arctangent function is 3sin(4x), not just x. This is where the chain rule comes into play. We differentiate the arctangent function with respect to its argument, 3sin(4x), and then multiply by the derivative of 3sin(4x) with respect to x. Applying the derivative of arctan, we get 1 / (1 + (3sin(4x))^2), which simplifies to 1 / (1 + 9sin^2(4x)). This is the first part of our derivative, but we're not done yet. We must now multiply by the derivative of the inner function, 3sin(4x). Now, let's find the derivative of 3sin(4x). Again, we encounter a composite function, this time the sine function with an argument of 4x. We apply the constant multiple rule and focus on differentiating sin(4x). The derivative of sin(x) is cos(x), so the derivative of sin(4x) will involve cos(4x). However, we must also account for the chain rule once more. We differentiate sin(4x) with respect to 4x, obtaining cos(4x), and then multiply by the derivative of 4x with respect to x, which is simply 4. Therefore, the derivative of sin(4x) is 4cos(4x), and the derivative of 3sin(4x) is 3 * 4cos(4x) = 12cos(4x). Finally, we assemble all the pieces. We have the constant multiple of 2, the derivative of the arctangent function, and the derivative of the inner sine function. Multiplying these together, we get: f'(x) = 2 * [1 / (1 + 9sin^2(4x))] * 12cos(4x). This is the derivative of f(x) = 2 arctan(3sin(4x)). We can further simplify this expression by combining the constants and writing it as: f'(x) = 24cos(4x) / (1 + 9sin^2(4x)). And there we have it! We have successfully navigated the intricacies of this composite function and found its derivative. The step-by-step approach, guided by the chain rule and the derivatives of basic functions, has revealed the solution in a clear and methodical manner. This process not only provides the answer but also deepens our understanding of the power and elegance of differential calculus.

The Derivative: f'(x) = 24cos(4x) / (1 + 9sin²(4x))

After a meticulous step-by-step process, applying the chain rule and the derivatives of trigonometric and inverse trigonometric functions, we have arrived at the derivative of f(x) = 2 arctan(3sin(4x)). The derivative, denoted as f'(x), is given by the expression: f'(x) = 24cos(4x) / (1 + 9sin²(4x)). This expression encapsulates the rate of change of the original function f(x) with respect to its input x. It provides valuable insights into the behavior of the function, such as its increasing and decreasing intervals, its critical points, and its concavity. Let's take a closer look at this derivative and discuss its implications. The derivative f'(x) is a quotient of two expressions, a numerator and a denominator. The numerator is 24cos(4x), which involves the cosine function. The cosine function oscillates between -1 and 1, so the numerator will oscillate between -24 and 24. The denominator is 1 + 9sin²(4x), which involves the square of the sine function. Since the square of any real number is non-negative, sin²(4x) will always be greater than or equal to 0. Therefore, 9sin²(4x) will also be non-negative, and 1 + 9sin²(4x) will always be greater than or equal to 1. This means that the denominator is always positive. Since the denominator is always positive and the numerator oscillates between -24 and 24, the derivative f'(x) will also oscillate. This indicates that the original function f(x) is neither always increasing nor always decreasing. It has intervals where it increases and intervals where it decreases. To find the intervals where f(x) is increasing or decreasing, we would need to analyze the sign of f'(x). f(x) is increasing where f'(x) is positive and decreasing where f'(x) is negative. The points where f'(x) changes sign are critical points of f(x). These points are potential locations of local maxima and minima. Furthermore, we can use the derivative f'(x) to find the second derivative, f''(x), which would provide information about the concavity of f(x). The second derivative tells us whether the function is concave up (shaped like a U) or concave down (shaped like an upside-down U). Points where the concavity changes are called inflection points. In summary, the derivative f'(x) = 24cos(4x) / (1 + 9sin²(4x)) is a powerful tool for understanding the behavior of the original function f(x) = 2 arctan(3sin(4x)). It allows us to analyze the function's rate of change, its increasing and decreasing intervals, its critical points, and its concavity. By exploring the derivative, we gain a deeper appreciation for the intricacies of the function and its properties.

Applications of Derivatives in Mathematics and Beyond

The derivative, as we've seen in the context of f(x) = 2 arctan(3sin(4x)), is not just an abstract mathematical concept; it is a powerful tool with a wide range of applications in various fields. Understanding the derivative and its applications is crucial for anyone pursuing studies or careers in mathematics, science, engineering, economics, and many other disciplines. In this section, we will explore some of the key applications of derivatives, highlighting their significance in solving real-world problems. One of the most fundamental applications of derivatives is in finding the slope of a tangent line to a curve at a given point. The derivative of a function at a specific point represents the instantaneous rate of change of the function at that point, which is geometrically equivalent to the slope of the tangent line. This concept is essential in understanding the behavior of curves and is used extensively in geometry and calculus. Another crucial application is in optimization problems. Derivatives help us find the maximum and minimum values of a function. This is incredibly useful in various fields, such as engineering, where we might want to design a structure that minimizes material usage while maximizing strength, or in economics, where we might want to maximize profit or minimize cost. The critical points of a function, where the derivative is zero or undefined, are the potential locations of these maximum and minimum values. In physics, derivatives are indispensable for describing motion and rates of change. Velocity, which is the rate of change of position with respect to time, is the first derivative of the position function. Acceleration, which is the rate of change of velocity with respect to time, is the second derivative of the position function. These concepts are fundamental to understanding the motion of objects and are used in mechanics, kinematics, and dynamics. In economics, derivatives are used to analyze marginal concepts. Marginal cost, marginal revenue, and marginal profit are all derivatives that represent the change in cost, revenue, and profit, respectively, resulting from a small change in production or sales. These concepts are crucial for making informed business decisions. Derivatives also play a vital role in related rates problems. These problems involve finding the rate of change of one quantity in terms of the rate of change of another related quantity. 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. In addition to these specific applications, derivatives are used extensively in mathematical modeling, numerical analysis, and differential equations. They are fundamental tools for understanding and solving problems in a wide range of scientific and engineering disciplines. In conclusion, the derivative is a versatile and powerful tool with far-reaching applications. From finding tangent lines and optimization to describing motion and analyzing marginal concepts, derivatives provide valuable insights and solutions to problems in mathematics and beyond. A solid understanding of derivatives is essential for anyone seeking to apply mathematical principles to real-world challenges.

Conclusion

In conclusion, our exploration of the derivative of f(x) = 2 arctan(3sin(4x)) has been a journey through the heart of differential calculus. We have successfully navigated the complexities of this composite function, employing the chain rule and the derivatives of trigonometric and inverse trigonometric functions to arrive at the solution: f'(x) = 24cos(4x) / (1 + 9sin²(4x)). This process has not only provided us with a concrete answer but has also reinforced our understanding of the fundamental principles of differentiation. We began by revisiting the essential concepts, including the chain rule, the derivatives of basic trigonometric functions, and the derivative of the arctangent function. These concepts served as our compass and map, guiding us through the intricacies of the problem. We then embarked on a step-by-step differentiation process, meticulously applying the chain rule to peel away the layers of the function. This methodical approach allowed us to break down the problem into manageable parts, ensuring clarity and accuracy. Along the way, we emphasized the importance of understanding the structure of the function and recognizing the presence of composite functions. Once we obtained the derivative, we took a moment to analyze its properties and implications. We discussed how the derivative can be used to determine the increasing and decreasing intervals of the original function, as well as its critical points and concavity. This analysis highlighted the power of the derivative as a tool for understanding the behavior of functions. Finally, we broadened our perspective by exploring the wide range of applications of derivatives in mathematics and beyond. We saw how derivatives are used in finding tangent lines, optimization problems, physics, economics, and various other fields. This exploration underscored the practical significance of differential calculus and its relevance to real-world challenges. The journey of finding the derivative of f(x) = 2 arctan(3sin(4x)) has been more than just a mathematical exercise. It has been an opportunity to deepen our understanding of calculus, to appreciate the elegance and power of differentiation, and to recognize the far-reaching applications of this fundamental concept. As we move forward in our mathematical endeavors, the lessons learned from this exploration will undoubtedly serve us well.