Find F Prime (π/3) For F(x) = 9 Sin X + Sin 3x
Introduction
In the realm of calculus, the derivative of a function holds immense significance. It provides insights into the rate of change of the function, revealing its behavior and characteristics. In this article, we delve into the fascinating world of derivatives by exploring the function f(x) = 9 sin x + sin 3x. Our primary objective is to determine the value of the derivative of this function, denoted as f'(x), at the specific point x = π/3. This exploration will not only enhance our understanding of derivatives but also showcase the application of differentiation rules in a practical scenario.
Understanding Derivatives: The Essence of Change
At its core, a derivative represents the instantaneous rate of change of a function. Imagine a curve plotted on a graph; the derivative at any point on this curve gives us the slope of the tangent line at that point. This slope signifies how the function's value is changing with respect to its input variable. Derivatives are instrumental in various fields, including physics, engineering, economics, and computer science, where understanding rates of change is crucial.
The Function at Hand: f(x) = 9 sin x + sin 3x
Our function of interest, f(x) = 9 sin x + sin 3x, is a trigonometric function comprising two sine terms with different frequencies. The term 9 sin x represents a sine wave with an amplitude of 9 and a frequency of 1, while sin 3x represents a sine wave with an amplitude of 1 and a frequency of 3. The combination of these two sine waves creates a more complex oscillatory pattern. To find the derivative of this function, we will employ the rules of differentiation, particularly the chain rule, which is essential when dealing with composite functions like sin 3x.
Navigating the Differentiation Process
To find f'(π/3), we must first determine the general expression for f'(x). This involves applying the rules of differentiation to the given function f(x). The derivative of sin x is cos x, and the derivative of sin 3x requires the chain rule. The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x). In our case, g(x) = 3x, so we need to multiply the derivative of the outer sine function by the derivative of the inner 3x term. Once we have found f'(x), we can substitute x = π/3 into the expression to obtain the desired value.
Step-by-Step Differentiation
Let's embark on the journey of finding the derivative of our function, f(x) = 9 sin x + sin 3x. We will break down the process into manageable steps, ensuring clarity and accuracy.
Step 1: Applying the Sum Rule
The first step in differentiating f(x) is to apply the sum rule. The sum rule states that the derivative of a sum of functions is the sum of their derivatives. In mathematical notation, this is expressed as (u + v)' = u' + v'. Applying this rule to our function, we get:
f'(x) = (9 sin x)' + (sin 3x)'
This step simplifies the problem by allowing us to differentiate each term separately.
Step 2: Differentiating 9 sin x
The next step is to differentiate the term 9 sin x. Here, we use the constant multiple rule, which states that the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function. Mathematically, this is represented as (cu)' = cu'. Applying this rule, we get:
(9 sin x)' = 9 (sin x)'
The derivative of sin x is cos x, a fundamental result in calculus. Therefore,
9 (sin x)' = 9 cos x
Step 3: Differentiating sin 3x
Differentiating sin 3x requires a bit more finesse due to the presence of the composite function. We employ the chain rule, which, as mentioned earlier, states that the derivative of f(g(x)) is f'(g(x)) * g'(x). In this case, f(u) = sin u and g(x) = 3x. The derivative of sin u with respect to u is cos u, and the derivative of 3x with respect to x is 3. Applying the chain rule, we get:
(sin 3x)' = cos(3x) * (3x)' = cos(3x) * 3 = 3 cos 3x
Step 4: Combining the Results
Now that we have differentiated each term separately, we combine the results to obtain the derivative of the entire function:
f'(x) = 9 cos x + 3 cos 3x
This expression gives us the derivative of f(x) for any value of x. Our next step is to evaluate this derivative at the specific point x = π/3.
Evaluating f'(π/3)
With the general expression for f'(x) in hand, we can now substitute x = π/3 to find the value of the derivative at this particular point. This will give us the slope of the tangent line to the curve of f(x) at x = π/3.
Step 1: Substituting x = π/3 into f'(x)
We substitute x = π/3 into the expression f'(x) = 9 cos x + 3 cos 3x:
f'(π/3) = 9 cos(π/3) + 3 cos(3 * π/3) = 9 cos(π/3) + 3 cos(π)
Step 2: Evaluating cos(π/3) and cos(π)
To proceed, we need to know the values of cos(π/3) and cos(π). Recall the unit circle and the special angles:
- cos(π/3) = 1/2
- cos(π) = -1
Step 3: Plugging in the Values
Substitute these values into the expression for f'(π/3):
f'(π/3) = 9 (1/2) + 3 (-1) = 9/2 - 3
Step 4: Simplifying the Expression
To simplify, we find a common denominator:
f'(π/3) = 9/2 - 6/2 = 3/2
Therefore, the value of the derivative of f(x) = 9 sin x + sin 3x at x = π/3 is 3/2.
Conclusion
In this article, we have successfully navigated the process of finding the derivative of the function f(x) = 9 sin x + sin 3x and evaluating it at the point x = π/3. We began by understanding the significance of derivatives as measures of instantaneous rates of change. We then applied the sum rule, constant multiple rule, and the chain rule to find the general expression for f'(x). Finally, we substituted x = π/3 into f'(x) and simplified the expression to obtain the result f'(π/3) = 3/2. This exercise highlights the power and elegance of calculus in analyzing and understanding the behavior of functions. The ability to find derivatives is a fundamental skill in mathematics and has wide-ranging applications in various scientific and engineering disciplines. Understanding these concepts deeply not only enhances mathematical prowess but also provides a valuable tool for problem-solving in diverse real-world scenarios.
This exploration underscores the importance of mastering differentiation techniques and their applications. As we continue our mathematical journey, the concepts and skills learned here will serve as a solid foundation for tackling more complex problems and exploring deeper mathematical territories. The world of calculus is vast and fascinating, and the journey of discovery is an ongoing and rewarding endeavor.
