Finding The Derivative Of F(x) = 7x^4√x + 3/(x^3√x)

Introduction: Unveiling the Secrets of Derivatives

Hey there, math enthusiasts! Today, we're going to tackle a fascinating problem involving derivatives. We've got a function, f(x) = 7x⁴√x + 3/(x³√x), that looks a bit intimidating at first glance. But don't worry, we'll break it down step by step and find its derivative, f'(x). This exploration will not only help us solve this specific problem but also enhance our understanding of derivative rules and how they apply to complex functions. Understanding derivatives is crucial in calculus, as they provide insights into the rate of change of a function. This concept has wide-ranging applications in physics, engineering, economics, and computer science. So, let’s roll up our sleeves and get started!

Before we jump into the calculations, let's quickly review the fundamental rules of differentiation. The power rule, the sum/difference rule, and the constant multiple rule are our primary tools for this task. The power rule states that the derivative of xⁿ is nxⁿ⁻¹. The sum/difference rule allows us to differentiate terms added or subtracted separately. The constant multiple rule states that the derivative of cf(x) is cf'(x), where c is a constant. Remember these, guys, because they're the bread and butter of differentiation. Now, with these rules in mind, we're well-equipped to tackle the given function and find its derivative. This methodical approach will not only lead us to the correct answer but also reinforce our understanding of these essential calculus concepts.

When dealing with functions like this, especially those involving radicals and fractions, rewriting them in a more manageable form is the first key step. We need to convert the square roots and fractions into exponential notation. Remember that √x can be written as x¹/², and 1/xⁿ can be written as x⁻ⁿ. Rewriting the function in this way makes it much easier to apply the power rule later on. So, let's take f(x) = 7x⁴√x + 3/(x³√x) and transform it into a form that’s easier to differentiate. This initial transformation is not just about making the function look neater; it's about setting the stage for a smoother and more accurate differentiation process. By rewriting the function in terms of exponents, we can avoid the pitfalls of directly applying the derivative rules to radicals and fractions, ensuring that we don’t make any unnecessary errors.

Step-by-Step Solution: Finding f'(x)

1. Rewriting the Function: Exponential Form

First, let's rewrite the function f(x) = 7x⁴√x + 3/(x³√x) using exponents. We know that √x = x¹/², so we can rewrite the first term as 7x⁴ * x¹/². For the second term, we have 3/(x³ * x¹/²). Now, let’s simplify these expressions. When multiplying terms with the same base, we add the exponents. For the first term, we have x⁴ * x¹/² = x⁴ + ¹/² = x⁹/². So the first term becomes 7x⁹/². For the second term, we have x³ * x¹/² = x³ + ¹/² = x⁷/². Therefore, the second term is 3/(x⁷/²), which can be rewritten as 3x⁻⁷/². Now our function looks like this: f(x) = 7x⁹/² + 3x⁻⁷/². See? Much simpler and ready for differentiation!

By converting the function to this form, we’ve made it significantly easier to apply the power rule, which is our next step. This process highlights the importance of algebraic manipulation in calculus. Often, the key to solving a calculus problem lies in simplifying the expression first. This step-by-step approach ensures that we are not only finding the derivative correctly but also reinforcing our understanding of mathematical principles. Remember, math is like building blocks—each step lays the foundation for the next. With the function now in its exponential form, we’re well-prepared to find its derivative with confidence and precision.

2. Applying the Power Rule

Now that we have f(x) = 7x⁹/² + 3x⁻⁷/², we can apply the power rule to each term. Remember, the power rule states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. Let’s apply this to the first term, 7x⁹/². Here, a = 7 and n = 9/2. So, the derivative of 7x⁹/² is (9/2) * 7 * x^(⁹/² - 1), which simplifies to (63/2)x⁷/². Now, let’s tackle the second term, 3x⁻⁷/². Here, a = 3 and n = -7/2. Applying the power rule, the derivative of 3x⁻⁷/² is (-7/2) * 3 * x^(-⁷/² - 1), which simplifies to (-21/2)x⁻⁹/².

The power rule is a cornerstone of differential calculus, and its correct application is essential for finding derivatives. By breaking down the function into its individual terms and applying the rule to each, we avoid common mistakes and ensure an accurate result. This methodical approach not only yields the correct derivative but also enhances our grasp of the underlying principles. Remember, practice makes perfect! The more we apply these rules, the more comfortable and proficient we become in using them. Now that we’ve differentiated each term separately, we’re ready to combine them to find the final form of f'(x).

3. Combining the Terms: The Final Derivative

We found the derivative of the first term to be (63/2)x⁷/² and the derivative of the second term to be (-21/2)x⁻⁹/². Now, we simply add these terms together to find f'(x). So, f'(x) = (63/2)x⁷/² + (-21/2)x⁻⁹/². We can rewrite this as f'(x) = (63/2)x⁷/² - (21/2)x⁻⁹/². This is the derivative of the given function. However, we can also rewrite this in a more readable form by converting the exponents back to radicals and fractions, if desired.

This step highlights the importance of algebraic manipulation in presenting the final result. While f'(x) = (63/2)x⁷/² - (21/2)x⁻⁹/² is mathematically correct, it's often helpful to express the derivative in a form that is easier to understand and interpret. Converting back to radicals and fractions can make the result more intuitive, especially in applied contexts. The ability to move fluidly between different forms of expressions is a valuable skill in calculus. With f'(x) now found, we've successfully navigated a complex problem using the fundamental principles of differentiation.

Alternative Representations and Simplifications

We've found f'(x) = (63/2)x⁷/² - (21/2)x⁻⁹/², but let's explore some alternative ways to represent this derivative. We can rewrite x⁷/² as x³√x and x⁻⁹/² as 1/(x⁴√x). Substituting these back into our expression, we get f'(x) = (63/2)x³√x - (21/2)(1/(x⁴√x)). Another way to simplify this expression is to find a common denominator. The common denominator for the two terms is 2x⁴√x. Multiplying the first term by x⁴√x / x⁴√x, we get [(63/2)x³√x * x⁴√x] / x⁴√x = (63/2)x⁷ / x⁴√x. The second term remains as -(21/2) / x⁴√x. Combining these, we have f'(x) = [(63/2)x⁷ - (21/2)] / x⁴√x. We can also factor out a 21/2 from the numerator to get f'(x) = (21/2)(3x⁷ - 1) / (x⁴√x). This simplified form can be more useful in certain applications.

Exploring different representations of the derivative is not just an academic exercise; it's a practical skill that can make further analysis easier. For example, finding the critical points of a function often requires setting its derivative equal to zero and solving for x. A simplified form of the derivative can make this process much more manageable. Moreover, understanding how to manipulate expressions algebraically is crucial for both theoretical and applied calculus. The ability to convert between different forms allows us to choose the representation that best suits the problem at hand. With this exploration, we’ve not only found the derivative but also honed our algebraic skills and gained a deeper appreciation for the flexibility of mathematical expressions.

Conclusion: Mastering the Art of Differentiation

So, we've successfully found the derivative of the function f(x) = 7x⁴√x + 3/(x³√x). We started by rewriting the function in exponential form, then applied the power rule, and finally, simplified the result. The derivative, f'(x), can be expressed in various forms, such as (63/2)x⁷/² - (21/2)x⁻⁹/² or (21/2)(3x⁷ - 1) / (x⁴√x). This exercise demonstrates the importance of understanding fundamental calculus rules and algebraic manipulation. By mastering these techniques, we can tackle even more complex problems with confidence.

Calculus, at its core, is about understanding change and rates of change. Derivatives are the key to unlocking this understanding. The journey we’ve taken today, from the initial problem to the final solution, is a microcosm of the broader process of mathematical problem-solving. It involves not only applying specific rules but also developing a strategic approach, simplifying expressions, and representing results in different ways. Remember, guys, that practice is essential! The more problems you solve, the more comfortable you'll become with these concepts. So, keep exploring, keep practicing, and keep pushing your mathematical boundaries. With a solid foundation in the fundamentals of calculus, you'll be well-equipped to tackle a wide range of challenges in mathematics and beyond.