Derivative Of G(x) = -2√(x)(x-1)²: A Step-by-Step Guide

Hey there, math enthusiasts! Ever stumbled upon a function that looks a bit intimidating and thought, "How do I even begin to find its derivative?" Well, today, we're going to tackle one such function head-on. We're diving deep into the world of calculus to explore the derivative of the function g(x) = -2√(x)(x-1)². This function might seem complex at first glance, but with a systematic approach and a sprinkle of calculus magic, we'll break it down step by step. So, buckle up and get ready to embark on this mathematical journey!

Understanding the Function: g(x) = -2√(x)(x-1)²

Before we jump into the differentiation process, let's take a moment to truly understand the function we're dealing with. Our function, g(x) = -2√(x)(x-1)², is a combination of several mathematical components. We have a square root term, √(x), which introduces a fractional exponent. We also have a squared term, (x-1)², which involves polynomial expansion. And, of course, we have a constant multiplier, -2, which simply scales the entire function. Recognizing these components is crucial because it helps us choose the appropriate differentiation rules and techniques. The square root function, often written as √x, can be expressed as x^(1/2). This representation is particularly useful when applying the power rule for differentiation. The term (x-1)² is a binomial squared, which can be expanded using the formula (a-b)² = a² - 2ab + b². This expansion will simplify the differentiation process later on. When we look at the function, we immediately notice the presence of both a product (between -2√(x) and (x-1)²) and composite functions (the square root and the squared term). This hints that we'll likely need to employ both the product rule and the chain rule during differentiation. Remember, the product rule states that the derivative of a product of two functions, u(x)v(x), is given by u'(x)v(x) + u(x)v'(x). The chain rule, on the other hand, helps us differentiate composite functions, and it states that the derivative of f(g(x)) is f'(g(x)) * g'(x). Keeping these rules in mind will be essential as we move forward. Now that we have a solid grasp of the function's structure, we're well-equipped to tackle its derivative. We'll proceed by carefully applying the product rule and the chain rule, breaking down the problem into manageable steps. Get ready to see how these calculus tools work together to unravel the derivative of g(x)!

Applying the Product Rule: The First Step

The first step in finding the derivative of g(x) is to recognize that it's a product of two functions: -2√(x) and (x-1)². This is where the product rule comes into play. The product rule, as we mentioned earlier, states that if we have a function that's the product of two other functions, say u(x) and v(x), then its derivative is given by: (uv)' = u'v + uv'. In our case, we can identify u(x) = -2√(x) and v(x) = (x-1)². Now, we need to find the derivatives of u(x) and v(x) separately. Let's start with u(x) = -2√(x). We can rewrite this as u(x) = -2x^(1/2). To find u'(x), we'll use the power rule, which states that the derivative of x^n is nx^(n-1). Applying the power rule, we get u'(x) = -2 * (1/2)x^(1/2 - 1) = -x^(-1/2). This can also be written as u'(x) = -1/√x. So, we've found the derivative of our first function! Next, let's tackle v(x) = (x-1)². To find v'(x), we'll use a combination of the power rule and the chain rule. We can think of v(x) as a composite function, where the outer function is squaring something and the inner function is (x-1). Applying the power rule and the chain rule, we get v'(x) = 2(x-1)^(2-1) * (1) = 2(x-1). So, we've also found the derivative of our second function. Now that we have u(x), v(x), u'(x), and v'(x), we can plug them into the product rule formula: g'(x) = u'(x)v(x) + u(x)v'(x). Substituting the values we found, we get: g'(x) = (-1/√x)(x-1)² + (-2√(x))(2(x-1)). This is the result of applying the product rule. However, we're not quite done yet! We need to simplify this expression to get a more manageable form. In the next section, we'll focus on simplifying this expression by combining terms and factoring out common factors. This will help us arrive at the final, most simplified form of the derivative.

Simplifying the Expression: Combining and Factoring

Alright, guys, we've made some good progress! We've successfully applied the product rule and found the initial form of g'(x). But as you can see, it's a bit messy and not exactly in its most user-friendly form. Our next mission is to simplify this expression. This involves combining terms and factoring out common factors to make the derivative look cleaner and easier to work with. Let's recap what we have so far: g'(x) = (-1/√x)(x-1)² + (-2√(x))(2(x-1)). The first thing we might notice is that both terms have a factor of (x-1). This is a great opportunity for factoring! We can factor out (x-1) from both terms, which gives us: g'(x) = (x-1)[(-1/√x)(x-1) + (-2√(x))(2)]. Now, the expression inside the brackets looks a bit more manageable. Let's focus on simplifying that further. We have two terms inside the brackets: (-1/√x)(x-1) and (-2√(x))(2). Let's simplify each of these individually. The first term, (-1/√x)(x-1), can be rewritten as -(x-1)/√x. The second term, (-2√(x))(2), simplifies to -4√(x). So, now we have: g'(x) = (x-1)[-(x-1)/√x - 4√(x)]. To combine the terms inside the brackets, we need to find a common denominator. In this case, the common denominator is √x. So, we'll rewrite -4√(x) as -4x/√x. Now we can combine the terms inside the brackets: g'(x) = (x-1)[-(x-1)/√x - 4x/√x] = (x-1)[(-x + 1 - 4x)/√x]. Simplifying the numerator inside the brackets, we get: g'(x) = (x-1)[(1 - 5x)/√x]. Finally, we can multiply the (x-1) term with the fraction inside the brackets: g'(x) = (x-1)(1 - 5x) / √x. And there you have it! We've successfully simplified the expression for g'(x). This form is much cleaner and easier to work with than our initial result after applying the product rule. However, we can still do a bit more to make it even more polished. In the next section, we'll explore further simplification techniques, such as rationalizing the denominator, to arrive at the final, most simplified form of the derivative.

Rationalizing the Denominator: The Final Polish

We've come a long way in finding and simplifying the derivative of our function! We've applied the product rule, combined terms, and factored out common factors. Now, we're in the final stretch: rationalizing the denominator. This is a common technique in calculus to eliminate square roots from the denominator of a fraction, making the expression look cleaner and more mathematically elegant. Let's remind ourselves of where we left off: g'(x) = (x-1)(1 - 5x) / √x. The denominator here is √x, which is an irrational number. To rationalize it, we need to multiply both the numerator and the denominator by √x. This will effectively eliminate the square root from the denominator. Multiplying the numerator and denominator by √x, we get: g'(x) = [(x-1)(1 - 5x) * √x] / [√x * √x]. The denominator simplifies to x, since √x * √x = x. So, we now have: g'(x) = [(x-1)(1 - 5x)√x] / x. We can further expand the numerator to get a more detailed expression. Let's expand (x-1)(1 - 5x): (x-1)(1 - 5x) = x - 5x² - 1 + 5x = -5x² + 6x - 1. So, our derivative becomes: g'(x) = [(-5x² + 6x - 1)√x] / x. This is a pretty clean and simplified form of the derivative! We've successfully rationalized the denominator and expanded the numerator to get a clear picture of the function's behavior. At this point, we've essentially reached the final form of the derivative. There might be other ways to express it, but this form is generally considered to be the most simplified and easy to interpret. To recap, we started with a seemingly complex function, g(x) = -2√(x)(x-1)², and we systematically applied the product rule, simplified the expression, and rationalized the denominator to arrive at its derivative: g'(x) = [(-5x² + 6x - 1)√x] / x. This journey showcases the power of calculus and how breaking down a problem into smaller, manageable steps can lead to a beautiful and elegant solution. In the next section, we'll summarize our findings and discuss the key takeaways from this exercise.

Conclusion: Key Takeaways and the Final Result

Wow, what a journey we've had! We started with a seemingly complex function, g(x) = -2√(x)(x-1)², and through careful application of calculus principles, we've successfully found its derivative. Let's take a moment to recap the key steps we took and the valuable lessons we learned along the way. First, we recognized the structure of the function, identifying it as a product of two functions: -2√(x) and (x-1)². This immediately hinted at the use of the product rule, which is a fundamental tool for differentiating products of functions. We then applied the product rule, which states that if g(x) = u(x)v(x), then g'(x) = u'(x)v(x) + u(x)v'(x). This gave us an initial expression for the derivative, but it was far from simplified. Next, we embarked on a simplification process. This involved factoring out common factors, combining like terms, and generally making the expression more manageable. Factoring out (x-1) was a key step in this process, as it significantly reduced the complexity of the expression. Finally, we rationalized the denominator. This is a common technique in calculus to eliminate square roots from the denominator, resulting in a cleaner and more mathematically elegant expression. We multiplied both the numerator and the denominator by √x, which successfully removed the square root from the denominator. Through these steps, we arrived at the final, simplified form of the derivative: g'(x) = [(-5x² + 6x - 1)√x] / x. This result represents the instantaneous rate of change of the function g(x) at any given point x. It's a powerful tool for understanding the behavior of the function, such as where it's increasing, decreasing, or has critical points. This exercise highlights the importance of a systematic approach to problem-solving in calculus. By breaking down a complex problem into smaller, manageable steps, we can apply the appropriate rules and techniques to arrive at a solution. It also reinforces the importance of simplification. A simplified expression is not only easier to work with, but it also provides a clearer understanding of the underlying mathematical relationships. So, the next time you encounter a seemingly daunting function, remember the steps we took today: identify the structure, apply the appropriate rules, simplify the expression, and don't be afraid to break it down into smaller pieces. With practice and a solid understanding of calculus principles, you'll be able to tackle even the most challenging derivatives! Thanks for joining me on this mathematical adventure. Keep exploring, keep learning, and keep those derivatives coming!