Derivative Of Y=(x⁵ Ln X)⁴ A Step-by-Step Solution
Calculating derivatives is a fundamental operation in calculus, essential for understanding rates of change and optimization problems. In this article, we will delve into the process of finding the derivative of a complex function: y = (x⁵ ln x)⁴. This problem combines the power rule, the product rule, and the chain rule, providing a rich example for mastering differentiation techniques. Our main focus will be on providing a detailed, step-by-step solution, ensuring clarity and comprehension for students and enthusiasts alike. We'll break down each step, explaining the underlying principles and how they apply to this specific problem. This will not only help you understand the solution but also equip you with the skills to tackle similar problems with confidence. Whether you are a student learning calculus or someone looking to refresh your knowledge, this guide will provide you with a solid understanding of the techniques involved in finding derivatives of complex functions. So, let's embark on this journey of mathematical exploration and unlock the secrets of differentiation.
Understanding the Problem
Before we dive into the solution, it's crucial to understand the structure of the function y = (x⁵ ln x)⁴. This function is a composition of several simpler functions, which means we'll need to use a combination of differentiation rules to find its derivative. The outermost function is a power function, raised to the power of 4, while the inner function is a product of a polynomial function (x⁵) and a logarithmic function (ln x). This intricate structure necessitates a methodical approach, applying the chain rule to address the outer function and the product rule to handle the inner function. In essence, we'll be peeling back the layers of the function, differentiating each part step-by-step. The key to success lies in recognizing these different layers and applying the appropriate rules in the correct order. This initial understanding sets the stage for a clear and efficient solution process, minimizing the chances of errors and maximizing comprehension. So, with this foundation in place, let's move on to the first step in our differentiation journey: applying the chain rule.
Applying the Chain Rule
The chain rule is a fundamental rule in calculus that allows us to differentiate composite functions, which are functions within functions. In our case, y = (x⁵ ln x)⁴, the outer function is u⁴, where u = x⁵ ln x is the inner function. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Mathematically, this is expressed as dy/dx = (dy/du) * (du/dx). Applying this to our problem, we first differentiate the outer function, treating the inner function as a single variable. The derivative of u⁴ with respect to u is 4u³. Next, we need to find the derivative of the inner function, x⁵ ln x, with respect to x. This is where the product rule comes into play, as we have a product of two functions. The chain rule provides the overarching framework for differentiating composite functions, ensuring that we account for the contribution of each layer in the function. By systematically applying this rule, we can break down complex differentiation problems into manageable steps, ultimately leading to the correct derivative. With the chain rule successfully applied, our next step is to tackle the derivative of the inner function using the product rule.
Utilizing the Product Rule
The product rule is another essential tool in calculus, used to find the derivative of a product of two functions. In our problem, the inner function u = x⁵ ln x is a product of two functions: f(x) = x⁵ and g(x) = ln x. The product rule states that the derivative of the product of two functions is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. Mathematically, this is expressed as d(fg)/dx = f'g + fg'. Applying this to our problem, we first find the derivatives of f(x) and g(x). The derivative of x⁵ is 5x⁴, using the power rule, and the derivative of ln x is 1/x. Now, we can apply the product rule: d(x⁵ ln x)/dx = (5x⁴)(ln x) + (x⁵)(1/x) = 5x⁴ ln x + x⁴. This result represents the derivative of the inner function, which we will later use in conjunction with the chain rule to find the overall derivative. The product rule allows us to systematically differentiate products of functions, ensuring that we account for the contribution of each factor. By mastering this rule, we can confidently tackle a wide range of differentiation problems. Having successfully applied the product rule, we are now ready to combine our results and find the final derivative.
Combining the Results and Simplifying
Now that we've applied both the chain rule and the product rule, it's time to combine our results and find the derivative of the original function, y = (x⁵ ln x)⁴. Recall that we found the derivative of the outer function (u⁴) with respect to u to be 4u³, and the derivative of the inner function (x⁵ ln x) with respect to x to be 5x⁴ ln x + x⁴. The chain rule tells us to multiply these two results together. So, dy/dx = 4(x⁵ ln x)³ * (5x⁴ ln x + x⁴). This is the derivative, but it's not yet in its simplest form. We can simplify this expression by factoring out x⁴ from the second term: dy/dx = 4(x⁵ ln x)³ * x⁴(5 ln x + 1). Further simplification involves combining the x terms: dy/dx = 4x¹⁵(ln x)³ * x⁴(5 ln x + 1) = 4x¹⁹(ln x)³(5 ln x + 1). This simplified expression represents the final derivative of the function. The process of combining results and simplifying is crucial in calculus, as it allows us to express the derivative in a clear and concise form. By carefully combining the results of the chain rule and the product rule, and then simplifying the expression, we have successfully found the derivative of the given function. With this final derivative in hand, let's reflect on the process and highlight the key takeaways.
Final Answer
After a detailed step-by-step process, we have successfully found the derivative of the function y = (x⁵ ln x)⁴. The final simplified derivative is: dy/dx = 4x¹⁹(ln x)³(5 ln x + 1). This solution showcases the power and versatility of calculus rules, particularly the chain rule and the product rule. By systematically applying these rules and simplifying the resulting expressions, we can tackle complex differentiation problems with confidence. Throughout this article, we emphasized the importance of understanding the structure of the function, breaking it down into its constituent parts, and applying the appropriate rules in the correct order. This methodical approach is key to success in calculus and beyond. We also highlighted the significance of simplifying the final result, ensuring that it is expressed in its most concise and understandable form. This final answer not only represents the solution to the problem but also serves as a testament to the power of calculus in solving real-world problems. Now, as we conclude this exploration of differentiation, let's recap the key takeaways and reinforce the concepts we've learned.
Key Takeaways
Throughout this exploration of finding the derivative of y = (x⁵ ln x)⁴, several key concepts and techniques have emerged. First and foremost, we've seen the importance of understanding the structure of a function before attempting to differentiate it. Recognizing the function as a composition of simpler functions, or as a product of functions, guides us in choosing the appropriate differentiation rules. The chain rule is crucial for differentiating composite functions, allowing us to peel back the layers and differentiate each part systematically. The product rule is essential for differentiating products of functions, ensuring that we account for the contribution of each factor. Furthermore, we've emphasized the importance of simplification. While the initial application of differentiation rules may yield a correct derivative, it's often necessary to simplify the expression to its most concise form. This not only makes the result easier to understand but also facilitates further calculations or analysis. Finally, we've highlighted the value of a methodical approach. By breaking down complex problems into smaller, manageable steps, we can minimize errors and maximize comprehension. These key takeaways serve as valuable lessons in calculus, applicable to a wide range of differentiation problems and beyond. By mastering these concepts and techniques, you'll be well-equipped to tackle any differentiation challenge that comes your way. As we conclude this guide, remember that practice is key to mastering calculus. So, keep exploring, keep differentiating, and keep expanding your mathematical horizons.
