Finding The Derivative Of X^(1/3)(1-3x)^4 A Step-by-Step Guide

In the realm of calculus, derivatives serve as a cornerstone, enabling us to unravel the rates at which functions change. Among the myriad functions encountered, expressions involving products and powers often present a unique challenge. This article delves into the intricate process of finding the derivative of the function f(x) = x^(1/3)(1-3x)4, offering a step-by-step guide to conquer this calculus hurdle. Understanding the derivative of such functions is crucial in various fields, including physics, engineering, and economics, where modeling dynamic systems and optimization problems often requires analyzing rates of change. This detailed exploration will not only equip you with the technical skills to solve similar problems but also enhance your conceptual grasp of differentiation techniques. We'll dissect the function, apply fundamental rules, and simplify the result to arrive at the derivative, ensuring clarity and accuracy at every step. Grasping this concept opens doors to more advanced calculus topics and real-world applications, making this a worthwhile endeavor for students and professionals alike.

Navigating the Product Rule: A Symphony of Functions

The function at hand, f(x) = x^(1/3)(1-3x)4, gracefully intertwines two distinct entities: x^(1/3) and (1-3x)^4. To unravel its derivative, we must invoke the product rule, a fundamental principle that orchestrates the differentiation of functions entwined in multiplication. This rule, a cornerstone of calculus, dictates that the derivative of a product is the sum of each function's derivative multiplied by the other function. Mathematically, this elegant principle is expressed as: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Here, u(x) and v(x) represent the individual functions that constitute the product, and the prime notation (') signifies their respective derivatives. Applying the product rule is not merely a mechanical process; it's a strategic maneuver that allows us to dissect a complex problem into more manageable parts. In our specific case, identifying x^(1/3) as u(x) and (1-3x)^4 as v(x) sets the stage for a systematic differentiation process. The product rule acts as a bridge, connecting the derivatives of the individual components to the derivative of the whole, enabling us to unravel the intricate dance of change within this composite function. This rule is not just a formula; it's a fundamental concept that underpins our understanding of how functions interact and evolve, making it indispensable in calculus and its applications.

Unveiling the Power Rule and Chain Rule: Essential Tools for Differentiation

Before we can fully apply the product rule to our function f(x) = x^(1/3)(1-3x)4, we must first equip ourselves with two more essential tools from the calculus arsenal: the power rule and the chain rule. The power rule is our guide when differentiating terms of the form x^n, where n is a constant. It elegantly states that the derivative of x^n is simply n*x^(n-1). This rule is fundamental for handling polynomial terms and fractional powers, like the x^(1/3) in our function. The chain rule, on the other hand, is our ally when differentiating composite functions – functions within functions. It dictates that the derivative of f(g(x)) is f'(g(x)) * g'(x). This rule becomes particularly crucial when dealing with expressions like (1-3x)^4, where a function (1-3x) is raised to a power. In our context, we'll employ the power rule to differentiate x^(1/3) and the chain rule to differentiate (1-3x)^4. The chain rule ensures we account for the inner function's derivative, providing a comprehensive understanding of how the entire composite function changes. These rules, working in tandem, empower us to dissect and differentiate complex expressions, paving the way for a complete solution to our derivative problem. Mastering these techniques is not just about memorizing formulas; it's about understanding the underlying principles of how functions transform and interact, a critical skill for any calculus practitioner.

Dissecting the Components: Differentiating x^(1/3) and (1-3x)^4

Now, let's apply the power rule and chain rule to the individual components of our function f(x) = x^(1/3)(1-3x)4. First, consider u(x) = x^(1/3). Applying the power rule, we bring the exponent (1/3) down and reduce the exponent by 1, resulting in u'(x) = (1/3)x^(1/3 - 1) = (1/3)x^(-2/3). This derivative represents the instantaneous rate of change of x^(1/3) with respect to x. Next, we tackle v(x) = (1-3x)^4. This requires the chain rule. We first differentiate the outer power function, treating (1-3x) as a single entity. This gives us 4(1-3x)^3. Then, we multiply by the derivative of the inner function (1-3x), which is -3. Combining these, we get v'(x) = 4(1-3x)^3 * (-3) = -12(1-3x)^3. This derivative captures the rate of change of (1-3x)^4, taking into account the interplay between the outer power function and the inner linear function. By dissecting the function into these manageable parts, we can apply the appropriate differentiation rules with precision. Each derivative, u'(x) and v'(x), provides a piece of the puzzle, contributing to the overall understanding of how f(x) changes. This meticulous approach ensures accuracy and lays the groundwork for the next step: applying the product rule to combine these derivatives.

Assembling the Puzzle: Applying the Product Rule

With the derivatives of our individual components, u'(x) = (1/3)x^(-2/3) and v'(x) = -12(1-3x)^3, in hand, we're now poised to apply the product rule. Recall that the product rule states: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Substituting our derivatives and original functions into this formula, we get: f'(x) = [(1/3)x^{(-2/3)][(1-3x)}4] + [x^{(1/3)][-12(1-3x)}3]. This expression represents the derivative of our original function, f(x) = x^(1/3)(1-3x)4. However, it's not yet in its most elegant form. The next step involves simplifying this expression, combining terms and factoring out common factors to arrive at a more concise and interpretable result. The product rule has served its purpose, guiding us through the differentiation process. Now, algebraic manipulation will refine our answer, revealing the underlying structure of the derivative and making it easier to analyze and apply in further calculations. This step is crucial not only for mathematical aesthetics but also for practical applications, as a simplified derivative is often easier to work with in problem-solving scenarios.

The Art of Simplification: Unveiling the Derivative's Essence

Our derivative, f'(x) = [(1/3)x^{(-2/3)][(1-3x)}4] + [x^{(1/3)][-12(1-3x)}3], while mathematically correct, is a bit unwieldy. To truly understand and utilize it effectively, we must embark on a journey of simplification. The key to this simplification lies in identifying and factoring out common factors. Observe that both terms contain factors of (1-3x) and powers of x. Specifically, we can factor out (1-3x)^3 and x^(-2/3). Factoring these terms out, we get: f'(x) = x^(-2/3)(1-3x)3 [(1/3)(1-3x) - 12x]. This step dramatically reduces the complexity of the expression, making it easier to manage and interpret. Next, we focus on simplifying the expression within the brackets. We distribute the (1/3) and combine like terms: (1/3)(1-3x) - 12x = (1/3) - x - 12x = (1/3) - 13x. Now, our derivative looks even cleaner: f'(x) = x^(-2/3)(1-3x)3 [(1/3) - 13x]. To further refine our result, we can eliminate the fraction within the brackets by multiplying the entire expression by 3/3: f'(x) = (1/3)x^(-2/3)(1-3x)3 [1 - 39x]. Finally, to express the derivative with positive exponents, we can rewrite x^(-2/3) as 1/x^(2/3). Our fully simplified derivative is: f'(x) = [(1-3x)^3 (1 - 39x)] / [3x^(2/3)]. This simplified form not only presents the derivative in a more elegant manner but also makes it easier to analyze its behavior, identify critical points, and apply it in various contexts. The art of simplification is not merely a cosmetic exercise; it's a crucial step in unlocking the derivative's true essence and making it a powerful tool for mathematical analysis.

The Grand Finale: The Simplified Derivative

After our meticulous journey through the realms of differentiation and simplification, we arrive at the grand finale: the simplified derivative of f(x) = x^(1/3)(1-3x)4. Our final answer, f'(x) = [(1-3x)^3 (1 - 39x)] / [3x^(2/3)], stands as a testament to the power of calculus and the elegance of mathematical manipulation. This expression encapsulates the instantaneous rate of change of the original function at any given point x. It allows us to understand how the function's output changes in response to infinitesimal changes in its input. The derivative is not just a formula; it's a window into the dynamic behavior of the function, revealing its increasing and decreasing intervals, its critical points, and its concavity. The process we've undertaken, from applying the product rule and chain rule to simplifying the expression, showcases the interconnectedness of mathematical concepts. Each step builds upon the previous one, culminating in a result that is both precise and insightful. This final derivative can now be used for a multitude of applications, from optimization problems to curve sketching to modeling physical phenomena. It is a tangible outcome of our intellectual endeavor, a tool that empowers us to analyze and understand the world around us through the lens of calculus. Mastering this process not only enhances our mathematical skills but also cultivates a deeper appreciation for the beauty and utility of calculus.

Keywords

Derivative, Product Rule, Chain Rule, Power Rule, Differentiation, Simplification, Calculus, Function, Rate of Change, Mathematical Analysis.