Demonstrating F'''(π/8) = 4√2 For F(x) = Cos(2x)

In the realm of calculus, derivatives play a pivotal role in understanding the rate of change of functions. Higher-order derivatives, in particular, provide insights into the concavity and inflection points of a function. This article delves into the fascinating world of derivatives by examining the function f(x) = cos(2x). Our primary goal is to demonstrate that the third derivative of this function, evaluated at x = π/8, yields the result 4√2. This exploration will not only showcase the application of differentiation rules but also highlight the elegance and precision of mathematical analysis. Let's embark on this journey by first understanding the function itself and then systematically calculating its derivatives.

Understanding the Function f(x) = cos(2x)

Before we dive into the derivatives, it's essential to grasp the nature of the function f(x) = cos(2x). This is a trigonometric function, specifically a cosine function, with a period that is compressed compared to the standard cosine function. The standard cosine function, cos(x), has a period of 2π, meaning it completes one full cycle over an interval of 2π. However, the 2x inside the cosine function compresses the period by a factor of 2. Therefore, f(x) = cos(2x) has a period of π. This means it oscillates twice as fast as the standard cosine function. Understanding this compression is crucial for visualizing the function's behavior and anticipating the behavior of its derivatives. The function oscillates between -1 and 1, just like the standard cosine function, but it does so more rapidly. This rapid oscillation will influence the magnitude of the derivatives, as we will see later.

The graph of f(x) = cos(2x) is a sinusoidal wave that oscillates around the x-axis. The amplitude of the wave is 1, meaning the maximum value is 1 and the minimum value is -1. The function starts at its maximum value (1) when x = 0, decreases to its minimum value (-1) at x = π/2, and returns to its maximum value at x = π. This cyclical behavior is fundamental to understanding the derivatives. The first derivative will describe the slope of this curve, the second derivative will describe the concavity, and the third derivative will describe the rate of change of the concavity. Visualizing the graph can provide valuable intuition about the signs and magnitudes of these derivatives at different points, especially at x = π/8, which is the point of interest in our problem.

Calculating the First Derivative: f'(x)

To find the first derivative, f'(x), we need to apply the chain rule. The chain rule is a fundamental rule in calculus that helps us differentiate composite functions, which are functions within functions. In our case, we have the cosine function acting on the function 2x. The chain rule states that the derivative of a composite function f(g(x)) is given by f'(g(x)) * g'(x). Applying this to f(x) = cos(2x), we first differentiate the outer function, which is cos(u), where u = 2x. The derivative of cos(u) with respect to u is -sin(u). Then, we differentiate the inner function, 2x, with respect to x, which gives us 2. Finally, we multiply these two results together. The chain rule is essential for correctly differentiating composite functions, and mastering it is crucial for calculus.

Therefore, the first derivative, f'(x), is given by:

f'(x) = -sin(2x) * 2 = -2sin(2x)

This first derivative represents the slope of the tangent line to the graph of f(x) = cos(2x) at any point x. The negative sign indicates that when sin(2x) is positive, the slope is negative, and vice versa. The factor of 2 in front of the sine function indicates that the slope changes more rapidly than it would for the standard sine function. Understanding the sign and magnitude of the first derivative is key to understanding where the original function is increasing or decreasing. At x = π/8, we can anticipate the value of f'(π/8) to be negative, as sin(2 * π/8) = sin(π/4) is a positive value.

Determining the Second Derivative: f''(x)

Now, let's move on to finding the second derivative, f''(x). The second derivative is the derivative of the first derivative, f'(x). So, we need to differentiate f'(x) = -2sin(2x) with respect to x. Again, we apply the chain rule. The outer function is -2sin(u), where u = 2x. The derivative of -2sin(u) with respect to u is -2cos(u). The derivative of the inner function, 2x, with respect to x is 2. Multiplying these two results together, we get:

f''(x) = -2cos(2x) * 2 = -4cos(2x)

The second derivative represents the concavity of the original function f(x) = cos(2x). A positive second derivative indicates that the function is concave up (shaped like a U), while a negative second derivative indicates that the function is concave down (shaped like an upside-down U). The second derivative is directly proportional to the original function, but with a negative sign and a factor of 4. This means that the concavity of f(x) is opposite in sign to the function itself and is scaled by a factor of 4. The second derivative provides critical information about the curvature of the function's graph.

Calculating the Third Derivative: f'''(x)

Finally, we arrive at the core of our problem: finding the third derivative, f'''(x). The third derivative is the derivative of the second derivative, f''(x). Therefore, we need to differentiate f''(x) = -4cos(2x) with respect to x. Once again, we employ the chain rule. The outer function is -4cos(u), where u = 2x. The derivative of -4cos(u) with respect to u is 4sin(u). The derivative of the inner function, 2x, with respect to x is 2. Multiplying these two results together, we obtain:

f'''(x) = 4sin(2x) * 2 = 8sin(2x)

The third derivative represents the rate of change of the concavity of the original function. It provides information about how the concavity is changing – whether it is becoming more concave up or more concave down. The third derivative, f'''(x) = 8sin(2x), is a sinusoidal function with an amplitude of 8 and the same period as f(x). This highlights a pattern: each derivative of a trigonometric function involves either a sine or cosine function, and the coefficients and signs alternate depending on the order of the derivative.

Evaluating the Third Derivative at x = π/8

Now that we have found the third derivative, f'''(x) = 8sin(2x), we can evaluate it at x = π/8. This means we substitute π/8 for x in the expression for f'''(x):

f'''(π/8) = 8sin(2 * π/8) = 8sin(π/4)

We know that sin(π/4) is equal to √2 / 2. Therefore,

f'''(π/8) = 8 * (√2 / 2) = 4√2

This confirms our initial goal: the third derivative of f(x) = cos(2x) evaluated at x = π/8 is indeed 4√2. This result demonstrates the power of calculus in providing precise numerical values for the rate of change of a function's concavity at a specific point. The fact that we obtained a positive value indicates that the concavity is increasing at x = π/8. This calculation encapsulates the entire process of finding higher-order derivatives and evaluating them at specific points.

Conclusion

In this exploration, we have successfully demonstrated that f'''(π/8) = 4√2 for the function f(x) = cos(2x). This involved a step-by-step process of calculating the first, second, and third derivatives using the chain rule and then evaluating the third derivative at the specified point. This exercise not only showcases the application of fundamental calculus principles but also highlights the intricate relationships between a function and its derivatives. The derivatives provide a deeper understanding of the function's behavior, including its slope, concavity, and the rate of change of concavity. This journey through derivatives illustrates the elegance and power of calculus in analyzing functions and their properties.

The process of finding higher-order derivatives is a cornerstone of calculus and has applications in various fields, including physics, engineering, and economics. For instance, in physics, the second derivative of position with respect to time represents acceleration, and the third derivative is sometimes referred to as jerk or jolt, representing the rate of change of acceleration. In engineering, derivatives are used to analyze the stability and behavior of systems. In economics, derivatives are used to model marginal costs and revenues. Mastering the concepts and techniques of differentiation is essential for anyone seeking to apply mathematical principles to real-world problems.

By understanding the derivatives of f(x) = cos(2x), we gain valuable insights into its behavior and its rate of change. The result f'''(π/8) = 4√2 is a testament to the precision and power of calculus in quantifying these changes. This exploration serves as a solid foundation for further delving into the world of calculus and its applications.