Finding The Derivative Of F(x)=(2x+8)^-3 And Evaluating F'(1)

Let's embark on a journey to find the derivative of the function f(x) = (2x + 8)^-3. This exploration will not only unveil the derivative, denoted as f'(x), but also provide a step-by-step understanding of the underlying principles of calculus. Derivatives play a pivotal role in various fields, including physics, engineering, and economics, as they represent the instantaneous rate of change of a function. In this context, we will leverage the chain rule, a fundamental concept in calculus, to unravel the derivative of our given function. This method allows us to tackle composite functions, where one function is nested within another. Understanding and applying the chain rule is essential for mastering differentiation and its applications.

To effectively find the derivative of f(x) = (2x + 8)^-3, we will employ the chain rule. This rule is a cornerstone of calculus, specifically designed for differentiating composite functions. Composite functions are essentially functions within functions, and the chain rule provides a systematic way to unravel their derivatives. In simpler terms, if we have a function y = f(g(x)), the chain rule dictates that the derivative dy/dx is given by the product of the derivative of the outer function f evaluated at g(x) and the derivative of the inner function g(x). Mathematically, this is expressed as dy/dx = f'(g(x)) * g'(x). This rule ensures that we account for the rate of change of both the outer and inner functions in determining the overall derivative. For our specific function, f(x) = (2x + 8)^-3, we can identify the outer function as u^-3 and the inner function as g(x) = 2x + 8. By applying the chain rule, we can systematically differentiate each part and combine the results to find the complete derivative.

First, we identify the outer and inner functions. The outer function is u^-3, where u represents the inner function. The inner function, in this case, is g(x) = 2x + 8. Now, we differentiate each function separately. The derivative of the outer function, u^-3, with respect to u, is -3u^-4. This is a straightforward application of the power rule, which states that the derivative of x^n is nx^(n-1). The derivative of the inner function, g(x) = 2x + 8, with respect to x, is simply 2. This follows from the power rule and the constant multiple rule, which state that the derivative of cx is c, where c is a constant. With the derivatives of both the outer and inner functions in hand, we can now apply the chain rule to find the derivative of the entire composite function.

Applying the chain rule, we multiply the derivative of the outer function by the derivative of the inner function. This gives us f'(x) = -3(2x + 8)^-4 * 2. This step is the heart of the chain rule, where we combine the individual rates of change to find the overall rate of change of the composite function. We are essentially accounting for how the outer function changes with respect to the inner function and how the inner function changes with respect to x. Now, we simplify the expression by multiplying the constants. This yields f'(x) = -6(2x + 8)^-4. This simplified expression represents the derivative of the original function, f(x) = (2x + 8)^-3. However, we can further refine this result by expressing it with a positive exponent. To do this, we simply move the term with the negative exponent to the denominator.

Finally, we rewrite the derivative with a positive exponent. The term (2x + 8)^-4 is equivalent to 1/(2x + 8)^4. Therefore, we can rewrite f'(x) as f'(x) = -6 / (2x + 8)^4. This is the final form of the derivative, expressing it in a clear and concise manner. This form is often preferred as it avoids negative exponents, making it easier to interpret and use in further calculations. The derivative f'(x) = -6 / (2x + 8)^4 tells us the instantaneous rate of change of the function f(x) = (2x + 8)^-3 at any given point x. This information is invaluable in various applications, such as optimization problems, curve sketching, and finding tangent lines.

Having determined the derivative f'(x) = -6 / (2x + 8)^4, our next task is to evaluate it at x = 1. This means we will substitute 1 for x in the expression for f'(x). Evaluating a derivative at a specific point provides us with the instantaneous rate of change of the function at that particular point. This is a crucial concept in calculus, as it allows us to understand the behavior of a function locally. In this case, finding f'(1) will tell us how the function f(x) = (2x + 8)^-3 is changing at the point where x = 1. This information can be used to approximate the function's value near x = 1, find the slope of the tangent line at that point, and gain insights into the function's increasing or decreasing behavior.

To find f'(1), we directly substitute x = 1 into the expression for the derivative, which is f'(x) = -6 / (2x + 8)^4. This substitution is a straightforward process, but it's crucial to perform it accurately to obtain the correct result. By replacing x with 1, we are essentially asking the derivative to tell us the instantaneous rate of change of the function at that specific point. This is analogous to zooming in on the graph of the function at x = 1 and examining the slope of the tangent line. The derivative at this point provides us with a precise measure of this slope, which is a key characteristic of the function's behavior at that location. The result of this substitution will be a numerical value representing the rate of change at x = 1.

Substituting x = 1 into the derivative, we get f'(1) = -6 / (2(1) + 8)^4. This expression now involves only arithmetic operations, which we can perform to simplify and obtain the final value. The order of operations is crucial here: we first perform the multiplication inside the parentheses, then the addition, then the exponentiation, and finally the division. Following this order ensures that we arrive at the correct result. The expression 2(1) + 8 simplifies to 2 + 8 = 10. This intermediate step is essential as it sets the base for the exponentiation. The subsequent calculation will involve raising this base to the power of 4, which will significantly affect the final value of the derivative.

Simplifying the expression, we have f'(1) = -6 / (10)^4. Now, we calculate 10^4, which is 10,000. This exponentiation step results in a large denominator, which will ultimately make the value of the derivative relatively small. This indicates that the function is changing slowly at x = 1. The derivative at a point provides a measure of the function's steepness at that point; a smaller value indicates a gentler slope, while a larger value indicates a steeper slope. In this case, the large denominator suggests that the function's graph is relatively flat near x = 1.

Therefore, f'(1) = -6 / 10000. This fraction can be further simplified. Dividing both the numerator and denominator by 2, we get f'(1) = -3 / 5000. This is the simplified fractional form of the derivative at x = 1. We can also express this as a decimal. Converting the fraction to a decimal, we find f'(1) = -0.0006. This decimal representation provides a clear sense of the magnitude of the derivative at x = 1. The negative sign indicates that the function is decreasing at this point, while the small absolute value (0.0006) confirms that the rate of decrease is quite slow. This means that as x increases slightly from 1, the function f(x) will decrease by a very small amount.

In conclusion, we have successfully found the derivative of f(x) = (2x + 8)^-3 to be f'(x) = -6 / (2x + 8)^4, and we have evaluated it at x = 1, obtaining f'(1) = -0.0006. This journey through differentiation has highlighted the power of the chain rule in handling composite functions and the significance of evaluating derivatives at specific points. The value of f'(1) provides crucial information about the function's behavior at x = 1, indicating that the function is decreasing at a very slow rate. This understanding of derivatives is fundamental in calculus and its applications, allowing us to analyze and predict the behavior of functions in various contexts.