Differentiating Y=x^2(x+1)(x^3+3x+1) A Step-by-Step Guide

Introduction

In this article, we will delve into the process of differentiating the function $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$. This problem involves the application of several important calculus concepts, including the product rule and the chain rule. Differentiation is a fundamental operation in calculus that allows us to find the rate of change of a function. In simpler terms, it helps us determine how a function's output changes as its input changes. The function we are dealing with, $y = x^2(x+1)(x^3+3x+1)$, is a product of three distinct terms: $x^2$, $(x+1)$, and $(x^3+3x+1)$. To differentiate such a product, we must employ the product rule. The product rule is a powerful tool in calculus that provides a method for differentiating the product of two or more functions. It states that the derivative of a product of functions is the sum of the derivatives of each function multiplied by the other functions. In this case, we have three functions multiplied together, so we will need to apply a generalized version of the product rule. Furthermore, the term $(x^3+3x+1)$ is a polynomial function, and differentiating it will require the power rule and the sum/difference rule. The power rule is a basic rule for differentiation that states that the derivative of $x^n$ is $nx^{n-1}$, where $n$ is any real number. The sum/difference rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives. By carefully applying these rules and concepts, we will be able to find the derivative of the given function and gain a deeper understanding of its behavior. The step-by-step approach will make the solution clear and easy to follow, even for those who are relatively new to calculus. So, let's embark on this journey of differentiation and unravel the intricacies of this problem.

Understanding the Product Rule

Before diving into the specific problem, it's crucial to have a solid understanding of the product rule. The product rule is a fundamental concept in calculus used to differentiate the product of two or more functions. Specifically, if we have two functions, $u(x)$ and $v(x)$, the derivative of their product, denoted as $(uv)'$ or $\frac{d}{dx}(u(x)v(x))$, is given by: $(uv)' = u'v + uv'$ This formula states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. This rule is essential when dealing with functions that are expressed as products, as it allows us to break down the differentiation process into smaller, more manageable steps. For the case of three functions, say $u(x)$, $v(x)$, and $w(x)$, the product rule can be extended as follows: $(uvw)' = u'vw + uv'w + uvw'$ This extended version tells us that the derivative of the product of three functions is the sum of three terms. Each term is formed by taking the derivative of one function and multiplying it by the other two functions. This pattern can be generalized to any number of functions, but for our problem, this three-function version is what we need. The product rule is not just a formula to memorize; it is a logical consequence of the definition of the derivative and the properties of limits. Understanding the underlying principle of the product rule can help in remembering and applying it correctly. For instance, one way to think about it is that when we have a product of functions, each function's change contributes to the overall change of the product. The product rule quantifies how these individual changes combine to give the total rate of change. In the context of our problem, $y = x^2(x+1)(x^3+3x+1)$, we can identify $u(x) = x^2$, $v(x) = (x+1)$, and $w(x) = (x^3+3x+1)$. Applying the extended product rule will involve finding the derivatives of each of these functions separately and then combining them in the appropriate manner. This approach allows us to systematically differentiate the complex product and arrive at the correct derivative. Therefore, a thorough understanding of the product rule is indispensable for solving this problem and many other differentiation problems in calculus. By grasping the concept and its application, we can confidently tackle functions expressed as products and unravel their rates of change.

Applying the Product Rule to the Given Function

Now, let's apply the product rule to our given function, $y = x^2(x+1)(x^3+3x+1)$. As we discussed, we can consider this function as a product of three functions: $u(x) = x^2$, $v(x) = (x+1)$, and $w(x) = (x^3+3x+1)$. To find the derivative of $y$ with respect to $x$, we will use the extended product rule, which states: $(uvw)' = u'vw + uv'w + uvw'$ First, we need to find the derivatives of each individual function. The derivative of $u(x) = x^2$ with respect to $x$ is found using the power rule, which states that the derivative of $x^n$ is $nx^{n-1}$. Applying this rule, we get: $u'(x) = \fracd}{dx}(x^2) = 2x$ Next, we find the derivative of $v(x) = (x+1)$ with respect to $x$. This is a simple linear function, and its derivative is $v'(x) = \frac{ddx}(x+1) = 1$ Finally, we need to find the derivative of $w(x) = (x^3+3x+1)$ with respect to $x$. This requires applying the power rule and the sum/difference rule. The power rule, as we mentioned, states that the derivative of $x^n$ is $nx^{n-1}$, and the sum/difference rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives. Applying these rules, we get $w'(x) = \frac{ddx}(x^3+3x+1) = 3x^2 + 3$ Now that we have the derivatives of each individual function, we can substitute them into the extended product rule formula $\frac{dy{dx} = u'vw + uv'w + uvw'$ $\frac{dy}{dx} = (2x)(x+1)(x^3+3x+1) + (x^2)(1)(x3+3x+1) + (x^2)(x+1)(3x2+3)$ This expression represents the derivative of $y$ with respect to $x$. However, it is in an unsimplified form. The next step involves expanding and simplifying this expression to obtain a more concise and manageable form of the derivative. This will involve careful algebraic manipulation, including distributing terms and combining like terms. While the expression we have obtained is technically the derivative, simplifying it will provide a clearer understanding of the function's rate of change and make it easier to work with in further calculations or applications. Therefore, the application of the product rule is just the first step in finding the derivative of this complex function; simplification is a crucial next step.

Expanding and Simplifying the Expression

After applying the product rule, we obtained the derivative of $y$ with respect to $x$ as: $\fracdy}{dx} = (2x)(x+1)(x^3+3x+1) + (x^2)(1)(x3+3x+1) + (x^2)(x+1)(3x2+3)$ Now, our task is to expand and simplify this expression. This process involves careful algebraic manipulation to combine like terms and obtain a more concise form of the derivative. Let's start by expanding each term individually. The first term is $(2x)(x+1)(x^3+3x+1)$. We can first multiply $(2x)(x+1)$ $(2x)(x+1) = 2x^2 + 2x$ Now, we multiply this result by $(x^3+3x+1)$: $(2x^2 + 2x)(x^3+3x+1) = 2x^5 + 6x^3 + 2x^2 + 2x^4 + 6x^2 + 2x$ $= 2x^5 + 2x^4 + 6x^3 + 8x^2 + 2x$ The second term is simply $(x^2)(1)(x^3+3x+1)$, which expands to: $(x^2)(x3+3x+1) = x^5 + 3x^3 + x^2$ The third term is $(x^2)(x+1)(3x^2+3)$. First, we multiply $(x^2)(x+1)$: $(x^2)(x+1) = x^3 + x^2$ Now, we multiply this result by $(3x^2+3)$: $(x^3 + x^2)(3x2+3) = 3x^5 + 3x^3 + 3x^4 + 3x^2$ Now, we add all three expanded terms together: $\frac{dydx} = (2x^5 + 2x^4 + 6x^3 + 8x^2 + 2x) + (x^5 + 3x^3 + x^2) + (3x^5 + 3x^4 + 3x^3 + 3x^2)$ Finally, we combine like terms $\frac{dy{dx} = (2x^5 + x^5 + 3x^5) + (2x^4 + 3x^4) + (6x^3 + 3x^3 + 3x^3) + (8x^2 + x^2 + 3x^2) + 2x$ $\frac{dy}{dx} = 6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$ This simplified expression represents the derivative of the given function $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$. The process of expanding and simplifying involved careful application of the distributive property and combining like terms. This step is crucial because the simplified form provides a clearer picture of the function's rate of change and is easier to use in further calculations or applications. The final expression, $6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$, is a polynomial function, and its coefficients reveal important information about the behavior of the original function. Thus, the simplification process is not just an algebraic exercise; it is a key step in understanding the derivative and its implications.

Final Result and Conclusion

After applying the product rule and carefully expanding and simplifying the expression, we have arrived at the derivative of the function $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$. The final result is: $\frac{dy}{dx} = 6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$ This polynomial function represents the instantaneous rate of change of the original function $y$ at any given point $x$. The derivative provides valuable information about the behavior of the original function, such as its increasing and decreasing intervals, critical points, and concavity. For instance, by setting the derivative equal to zero and solving for $x$, we can find the critical points of the original function, which are the points where the function's slope is zero. These points can be local maxima, local minima, or saddle points. The sign of the derivative in different intervals tells us whether the function is increasing or decreasing in those intervals. If the derivative is positive in an interval, the function is increasing, and if it is negative, the function is decreasing. Furthermore, we can find the second derivative of the function by differentiating the first derivative. The second derivative provides information about the concavity of the original function. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down. In conclusion, differentiating $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$ involved the application of the product rule, which is a fundamental tool in calculus for differentiating products of functions. The process also required careful algebraic manipulation to expand and simplify the resulting expression. The final result, $\frac{dy}{dx} = 6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$, provides a complete description of the function's rate of change and can be used to analyze its behavior in detail. This exercise highlights the power of calculus in understanding and characterizing functions, and it demonstrates the importance of mastering techniques like the product rule and algebraic simplification. The derivative we found is not just a formula; it is a powerful tool for understanding the dynamics of the original function.