Differentiating Y=x^2(x+1)(x^3+3x+1) With Respect To X A Calculus Guide

Introduction

In the realm of calculus, differentiation is a fundamental operation that allows us to determine the rate at which a function's output changes with respect to its input. In simpler terms, it helps us find the slope of a curve at any given point. This article delves into the process of differentiating the function $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$. This particular function is a product of three terms, which necessitates the use of the product rule of differentiation. The product rule is a powerful tool that enables us to differentiate functions that are expressed as the product of two or more terms. Mastering the product rule is crucial for tackling more complex differentiation problems. Understanding the steps involved in differentiating this function will not only enhance your calculus skills but also provide a solid foundation for solving more intricate problems in various fields, including physics, engineering, and economics. The key to success in differentiation lies in a methodical approach and a clear understanding of the underlying principles. We will break down the problem into manageable steps, ensuring that each step is thoroughly explained and justified. This approach will not only help you solve this particular problem but also equip you with the skills to tackle similar problems with confidence. Let's embark on this journey of differentiation and unravel the intricacies of this function.

Understanding the Product Rule

Before diving into the differentiation of our specific function, it's crucial to understand the product rule. The product rule is a fundamental concept in calculus that provides a method for differentiating functions that are expressed as the product of two or more terms. In essence, it states that the derivative of a product of functions is not simply the product of their derivatives. Instead, it involves a more nuanced approach that considers the interplay between the functions. Mathematically, if we have two functions, say $u(x)$ and $v(x)$, then the derivative of their product, denoted as $(uv)'$ or $\frac{d}{dx}(uv)$, is given by the following formula:

$$\frac{d}{dx}(uv) = u'v + uv'$$

Where $u'$ and $v'$ represent the derivatives of $u(x)$ and $v(x)$ with respect to $x$, respectively. This formula tells us that to differentiate the product of two functions, we need to take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function. This may seem a bit complex at first, but with practice, it becomes a straightforward and powerful technique. The product rule can be extended to products of more than two functions. For instance, if we have three functions, $u(x)$, $v(x)$, and $w(x)$, then the derivative of their product is given by:

$$\frac{d}{dx}(uvw) = u'vw + uv'w + uvw'$$

This pattern can be generalized to any number of functions. The key is to differentiate one function at a time while keeping the others constant, and then sum up all the resulting terms. Mastering the product rule is essential for differentiating complex functions and is a cornerstone of calculus. It's not just a mathematical formula; it's a tool that allows us to analyze and understand the behavior of functions in various contexts. In the following sections, we will apply this rule to differentiate the given function, breaking down the process into manageable steps and providing clear explanations along the way. Understanding the nuances of the product rule will not only help you solve this specific problem but also equip you with the skills to tackle a wide range of differentiation problems.

Applying the Product Rule to $y=x^2(x+1)(x^3+3x+1)$

Now, let's apply the product rule to differentiate the given function: $y = x^2(x+1)(x^3+3x+1)$. This function is a product of three terms: $x^2$, $(x+1)$, and $(x^3+3x+1)$. To make the differentiation process more manageable, we can initially treat the function as a product of two terms, say $u$ and $v$, where:

• $u = x^2(x+1)$
• $v = (x^3+3x+1)$

Then, we can rewrite the function as $y = uv$ and apply the product rule: $\frac{dy}{dx} = u'v + uv'$. First, we need to find the derivatives of $u$ and $v$ with respect to $x$. Let's start with $u = x^2(x+1)$. This term is itself a product of two functions, $x^2$ and $(x+1)$. So, we need to apply the product rule again. Let $p = x^2$ and $q = (x+1)$. Then, $u = pq$, and its derivative $u'$ is given by:

$$u' = p'q + pq'$$

The derivatives of $p$ and $q$ are:

• $p' = \frac{d}{dx}(x^2) = 2x$
• $q' = \frac{d}{dx}(x+1) = 1$

Substituting these into the expression for $u'$, we get:

$$u' = (2x)(x+1) + (x^2)(1) = 2x^2 + 2x + x^2 = 3x^2 + 2x$$

Now, let's find the derivative of $v = (x^3+3x+1)$. This is a straightforward application of the power rule and the sum rule of differentiation:

$$v' = \frac{d}{dx}(x^3+3x+1) = \frac{d}{dx}(x^3) + 3\frac{d}{dx}(x) + \frac{d}{dx}(1) = 3x^2 + 3$$

We have now found $u'$ and $v'$. We can substitute these into the product rule formula for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = u'v + uv' = (3x^2 + 2x)(x^3+3x+1) + (x^2(x+1))(3x^2 + 3)$$

This expression gives us the derivative of $y$ with respect to $x$. In the next section, we will simplify this expression to obtain a more concise form. The process of applying the product rule may seem lengthy, but it is a systematic way to differentiate complex functions. By breaking down the problem into smaller steps and carefully applying the rules of differentiation, we can arrive at the correct solution. The key is to be patient, methodical, and to double-check each step to avoid errors. The more you practice, the more comfortable you will become with these techniques, and the faster you will be able to solve differentiation problems.

Simplifying the Expression

Having obtained the derivative expression, our next step is to simplify it. The expression we derived in the previous section is:

$$\frac{dy}{dx} = (3x^2 + 2x)(x^3+3x+1) + (x^2(x+1))(3x^2 + 3)$$

To simplify this, we need to expand the products and combine like terms. Let's start by expanding the first product:

$$(3x^2 + 2x)(x^3+3x+1) = 3x^2(x^3+3x+1) + 2x(x^3+3x+1)$$

$$= 3x^5 + 9x^3 + 3x^2 + 2x^4 + 6x^2 + 2x$$

$$= 3x^5 + 2x^4 + 9x^3 + 9x^2 + 2x$$

Now, let's expand the second product:

$$(x^2(x+1))(3x^2 + 3) = x^2(x+1)(3x^2 + 3)$$

First, expand $x^2(x+1)$:

$$x^2(x+1) = x^3 + x^2$$

Then, multiply by $(3x^2 + 3)$:

$$(x^3 + x^2)(3x^2 + 3) = x^3(3x^2 + 3) + x^2(3x^2 + 3)$$

$$= 3x^5 + 3x^3 + 3x^4 + 3x^2$$

Now, we can add the two expanded expressions together:

$$\frac{dy}{dx} = (3x^5 + 2x^4 + 9x^3 + 9x^2 + 2x) + (3x^5 + 3x^4 + 3x^3 + 3x^2)$$

Combine like terms:

$$\frac{dy}{dx} = (3x^5 + 3x^5) + (2x^4 + 3x^4) + (9x^3 + 3x^3) + (9x^2 + 3x^2) + 2x$$

$$\frac{dy}{dx} = 6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$$

This is the simplified expression for the derivative of $y$ with respect to $x$. The process of simplification is crucial because it allows us to express the derivative in a more concise and manageable form. This not only makes it easier to work with the derivative in subsequent calculations but also provides a clearer understanding of the function's behavior. The simplified expression reveals the polynomial nature of the derivative and allows us to analyze its roots, critical points, and other important features. Moreover, a simplified expression is less prone to errors in further computations. Therefore, taking the time to simplify the derivative is a worthwhile investment in the overall problem-solving process. The techniques used here, such as expanding products and combining like terms, are fundamental algebraic skills that are essential for success in calculus and other areas of mathematics.

Final Answer

In conclusion, by applying the product rule and simplifying the resulting expression, we have found the derivative of $y = x^2(x+1)(x^3+3x+1)$ with respect to $x$. The final answer is:

$$\frac{dy}{dx} = 6x^5 + 5x^4 + 12x^3 + 12x^2 + 2x$$

This result represents the instantaneous rate of change of the function $y$ with respect to $x$. It provides valuable information about the function's behavior, such as its slope at any given point. The process of differentiating this function has demonstrated the power and versatility of the product rule, a fundamental tool in calculus. By breaking down the problem into manageable steps and carefully applying the rules of differentiation, we were able to arrive at the solution. This methodical approach is crucial for tackling more complex differentiation problems. The ability to differentiate functions is a cornerstone of calculus and has wide-ranging applications in various fields, including physics, engineering, economics, and computer science. Understanding the underlying principles and mastering the techniques of differentiation will equip you with the skills to solve a wide range of problems and gain a deeper understanding of the world around us. This exercise has not only provided a solution to a specific problem but has also reinforced the importance of a systematic approach, careful application of rules, and the value of simplification in mathematical problem-solving. The journey of differentiation is a rewarding one, and with practice and perseverance, you can master the art of finding derivatives and unlock the power of calculus.